Explanation:
Okay, so I need to compute the value of this expression: 1 over cos squared 10 degrees plus 1 over sin squared 20 degrees plus 1 over sin squared 40 degrees. Let me write that down to make sure I have it right:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ}
$$

Hmm, okay. So, first, I should recall some trigonometric identities that might help simplify this. The denominators are all squared trigonometric functions, so maybe I can use identities related to that. Let me think... I know that 1 over cos squared theta is secant squared theta, and similarly for sine. Also, there's the identity that 1 + tan squared theta equals secant squared theta. Maybe that can be useful here?

But before jumping into that, let me check if there's a pattern or some relationship between the angles. The angles given are 10°, 20°, and 40°. I notice that 10°, 20°, 40° are all multiples of 10°, and they seem to be doubling each time. Wait, 10°, then 20°, then 40°, which is doubling each time. Maybe there's some recursive relationship or something?

Alternatively, maybe I can convert all angles to radians? Although I don't know if that would help. Let me think. Alternatively, maybe using some angle addition formulas or product-to-sum formulas?

Alternatively, perhaps I can express all terms in terms of sine or cosine. Let me see. For example, cos 10° is related to sin 80°, since cos(theta) = sin(90° - theta). So cos 10° = sin 80°, so 1 over cos squared 10° is 1 over sin squared 80°. So maybe rewriting the first term as 1/sin²80°? Let me check:

Yes, cos(10°) = sin(80°), so 1/cos²(10°) = 1/sin²(80°). So then the expression becomes:

1/sin²(80°) + 1/sin²(20°) + 1/sin²(40°)

Hmm, so now all terms are in terms of sine squared. Maybe that's helpful? Not sure yet. Let me think about other identities.

Alternatively, maybe I can use the identity for 1/sin²(theta). Since 1/sin²(theta) is equal to 1 + cot²(theta). Wait, but that might not be helpful here. Alternatively, maybe using some other identities?

Alternatively, perhaps I can use the identity for sin(3θ) or something? Let me think. For example, sin(3θ) = 3 sinθ - 4 sin³θ. But I don't know if that helps here. Alternatively, maybe using product formulas?

Alternatively, maybe I can consider that 10°, 20°, 40° are related to angles that add up to certain values. For example, 10° + 20° + 40° = 70°, which doesn't seem particularly helpful. Alternatively, maybe considering complementary angles?

Alternatively, maybe I can use the identity for the sum of 1/sin²(theta) terms. Wait, but I don't recall a direct formula for that. Maybe I need to compute each term separately?

Alternatively, perhaps there's a clever substitution or a way to combine these terms. Let me think. Let me first compute each term numerically to get an idea of what the answer might be. Maybe that can help me check my work later.

Let me calculate each term:

First term: 1 / cos²(10°). Let me compute cos(10°). Cos(10°) is approximately cos(0.1745 radians). Let me recall that cos(0°) is 1, cos(30°) is about 0.866. So cos(10°) is about 0.9848. Therefore, cos²(10°) is approximately (0.9848)^2 ≈ 0.9698. Therefore, 1 / 0.9698 ≈ 1.031.

Second term: 1 / sin²(20°). Sin(20°) is approximately 0.3420, so sin²(20°) ≈ 0.11699, so 1 / 0.11699 ≈ 8.545.

Third term: 1 / sin²(40°). Sin(40°) is approximately 0.6428, so sin²(40°) ≈ 0.4132, so 1 / 0.4132 ≈ 2.420.

Adding them up: 1.031 + 8.545 + 2.420 ≈ 11.996. So approximately 12. So maybe the exact value is 12? That seems possible. Let me check with more accurate calculations.

Wait, let me use more precise values.

First, cos(10°):

cos(10°) ≈ 0.984807753012208

So cos²(10°) ≈ (0.984807753012208)^2 ≈ 0.969846310392847

Therefore, 1 / cos²(10°) ≈ 1.03109527272276

Second term: sin(20°) ≈ 0.3420201433256687

sin²(20°) ≈ (0.3420201433256687)^2 ≈ 0.116977778440511

Therefore, 1 / sin²(20°) ≈ 8.54582352624225

Third term: sin(40°) ≈ 0.6427876096865393

sin²(40°) ≈ (0.6427876096865393)^2 ≈ 0.413175911166488

Therefore, 1 / sin²(40°) ≈ 2.42034534307822

Adding them up:

1.03109527272276 + 8.54582352624225 + 2.42034534307822 ≈

First, 1.031095 + 8.545823 ≈ 9.576918, then + 2.420345 ≈ 11.997263. So approximately 12. So that suggests that the exact value is 12. Therefore, maybe the answer is 12? But I need to verify this analytically.

So, the problem is to show that:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} = 12
$$

So, how can I show this?

Let me think. Let me recall that 1/cos²θ = 1 + tan²θ, so maybe I can write each term as 1 + tan²θ, but I don't know if that helps. Alternatively, maybe using some identities for multiple angles?

Alternatively, perhaps using complex numbers or roots of unity? Not sure. Alternatively, maybe using the identity for sin(3θ) or something?

Wait, let me think about the angles. Let me note that 10°, 20°, 40°, and maybe 80°, which is related to the first term. Also, 10°, 20°, 40°, 80°, which are all multiples of 10°, and they are related to each other through doubling. Also, 10° * 3 = 30°, which is a standard angle. Maybe there's a connection here.

Alternatively, perhaps using the identity for sin(3θ). Let me recall that sin(3θ) = 3 sinθ - 4 sin³θ. Let me see if that can be used here. For example, if I take θ = 10°, then sin(30°) = 3 sin(10°) - 4 sin³(10°). Since sin(30°) is 1/2. So:

1/2 = 3 sin(10°) - 4 sin³(10°)

But I don't know if that helps here. Alternatively, maybe considering equations involving multiple angles?

Alternatively, maybe using the identity for 1/sin²θ. Let me recall that 1/sin²θ = 1 + cot²θ, but again, not sure.

Alternatively, maybe using the identity for the sum of 1/sin²θ terms. Wait, but I need to find a way to combine them. Let me think.

Alternatively, maybe using the identity for the sum of 1/sin²θ for specific angles. Let me think about the angles 10°, 20°, 40°. Let me note that 10°, 20°, 40°, 80°, etc., are related to each other through doubling. Also, 10°, 20°, 40°, 80°, and 160°, etc. Maybe there's a pattern here.

Alternatively, perhaps using the identity for the sum of 1/sin²θ over certain angles. Wait, but I need to think of a way to relate them.

Alternatively, maybe using the identity for tan(3θ). Let me recall that tan(3θ) = (3 tanθ - tan³θ) / (1 - 3 tan²θ). Maybe that can be useful?

Alternatively, let me think about writing all terms in terms of sine or cosine. Let me try to convert the first term to sine:

As I did before, 1/cos²(10°) = 1/sin²(80°). So the expression becomes:

1/sin²(80°) + 1/sin²(20°) + 1/sin²(40°)

Now, note that 80°, 20°, 40°, and maybe 60°, which is another angle. Let me think. Also, 80° is 60° + 20°, 40° is 60° - 20°, etc. Maybe using sum formulas?

Alternatively, maybe using the identity for sin(3θ). Let me think. For example, sin(3θ) = 3 sinθ - 4 sin³θ. Let me try to use this for θ = 20°, which gives sin(60°) = 3 sin(20°) - 4 sin³(20°). Since sin(60°) is √3/2 ≈ 0.8660. So:

√3/2 = 3 sin(20°) - 4 sin³(20°)

But again, not sure how that helps here. Alternatively, maybe solving for sin(20°), but that seems complicated.

Alternatively, maybe considering that 80°, 40°, 20° are related to each other. For example, 80° = 2*40°, 40° = 2*20°, so maybe using double-angle identities?

Let me recall that sin(2θ) = 2 sinθ cosθ. So, for example, sin(40°) = 2 sin(20°) cos(20°). Similarly, sin(80°) = 2 sin(40°) cos(40°). So maybe I can express sin(80°) in terms of sin(40°) and cos(40°), and so on.

But how does that help with 1/sin²(theta)?

Alternatively, maybe using the identity for 1/sin²(theta) in terms of cotangent or something else. Wait, 1/sin²(theta) is equal to 1 + cot²(theta). So, if I can express each term as 1 + cot²(theta), then maybe I can sum them up. Let me try that.

So, the original expression:

1/cos²(10°) + 1/sin²(20°) + 1/sin²(40°) = [1 + tan²(10°)] + [1 + cot²(20°)] + [1 + cot²(40°)] = 3 + tan²(10°) + cot²(20°) + cot²(40°)

Hmm, not sure if that helps. Maybe not directly. Let me think.

Alternatively, maybe using some other identities. Let me think about the first term, 1/cos²(10°). If I can relate this to other terms. For example, using the identity for tan(3θ). Let me try θ = 10°, then tan(30°) = tan(3*10°) = (3 tan10° - tan³10°)/(1 - 3 tan²10°). Since tan(30°) is 1/√3 ≈ 0.577. Let me denote t = tan(10°). Then:

(3t - t³)/(1 - 3t²) = 1/√3

Cross multiplying:

√3(3t - t³) = 1 - 3t²

Which gives:

3√3 t - √3 t³ = 1 - 3t²

Rearranged:

√3 t³ - 3t² - 3√3 t + 1 = 0

Hmm, this seems complicated. Maybe not helpful here.

Alternatively, maybe considering that 10°, 20°, 40°, and 80° are related to each other through complementary angles or supplementary angles. For example, 80° is complementary to 10°, 40° is complementary to 50°, but not sure.

Alternatively, maybe using product formulas. For example, the product of sines or cosines. But again, not sure.

Alternatively, maybe using the identity for the sum of 1/sin²(theta) terms. Wait, I remember that there is an identity for the sum of 1/sin²(theta) over certain angles. Wait, but I don't recall exactly. Maybe I need to derive it.

Alternatively, let me think about writing all terms with denominator sin²(theta) and trying to combine them. Let me note that 1/sin²(80°) + 1/sin²(20°) + 1/sin²(40°). Let me note that 80° = 60° + 20°, 40° = 60° - 20°, so maybe using sum formulas?

Alternatively, maybe using the identity for sin(A) + sin(B) or something else. Wait, but we have squares in the denominator. Maybe not.

Alternatively, maybe using the identity for 1/sin²(theta) = 2/(1 - cos(2 theta)). Let me recall that:

Since 1 - cos(2θ) = 2 sin²θ, so 1/sin²θ = 2 / (1 - cos(2θ)). Therefore, maybe I can rewrite each term as:

For the first term, 1/sin²(80°) = 2 / (1 - cos(160°))

Second term, 1/sin²(20°) = 2 / (1 - cos(40°))

Third term, 1/sin²(40°) = 2 / (1 - cos(80°))

Therefore, the entire expression becomes:

2 / (1 - cos(160°)) + 2 / (1 - cos(40°)) + 2 / (1 - cos(80°))

Hmm, so that's 2 times [1/(1 - cos160°) + 1/(1 - cos40°) + 1/(1 - cos80°)]

Now, let me compute these denominators. Let me note that cos(160°) is equal to cos(180° - 20°) = -cos(20°). Similarly, cos(80°) is cos(80°), and cos(40°) is cos(40°). Therefore:

1 - cos(160°) = 1 - (-cos20°) = 1 + cos20°

Similarly, 1 - cos(40°) remains as is, and 1 - cos(80°) remains as is.

Therefore, the expression becomes:

2 [ 1/(1 + cos20°) + 1/(1 - cos40°) + 1/(1 - cos80°) ]

Hmm, so now I have:

2 [ 1/(1 + cos20°) + 1/(1 - cos40°) + 1/(1 - cos80°) ]

Now, maybe I can simplify each term. Let me recall that 1/(1 ± cosθ) can be expressed using identities. For example, 1/(1 + cosθ) = (1 - cosθ)/sin²θ, and 1/(1 - cosθ) = (1 + cosθ)/sin²θ. Wait, let me check:

Yes, because:

Multiply numerator and denominator by (1 - cosθ) for 1/(1 + cosθ):

[1 * (1 - cosθ)] / [(1 + cosθ)(1 - cosθ)] = (1 - cosθ)/sin²θ

Similarly, for 1/(1 - cosθ):

Multiply numerator and denominator by (1 + cosθ):

[1 * (1 + cosθ)] / [(1 - cosθ)(1 + cosθ)] = (1 + cosθ)/sin²θ

Therefore, substituting back into the expression:

2 [ (1 - cos20°)/sin²20° + (1 + cos40°)/sin²40° + (1 + cos80°)/sin²80° ]

Wait, but let me check:

Original terms:

1/(1 + cos20°) = (1 - cos20°)/sin²20°

Similarly, 1/(1 - cos40°) = (1 + cos40°)/sin²40°

And 1/(1 - cos80°) = (1 + cos80°)/sin²80°

Therefore, the entire expression becomes:

2 [ (1 - cos20°)/sin²20° + (1 + cos40°)/sin²40° + (1 + cos80°)/sin²80° ]

Hmm, so now I have:

2 times [ (1 - cos20°)/sin²20° + (1 + cos40°)/sin²40° + (1 + cos80°)/sin²80° ]

Now, let me note that 1 - cos20° is 2 sin²10°, from the identity 1 - cosθ = 2 sin²(θ/2). Similarly, 1 + cosθ = 2 cos²(θ/2). Therefore:

Let me rewrite each term:

First term: (1 - cos20°)/sin²20° = [2 sin²10°] / sin²20°

Second term: (1 + cos40°)/sin²40° = [2 cos²20°] / sin²40°

Third term: (1 + cos80°)/sin²80° = [2 cos²40°] / sin²80°

Therefore, substituting back:

2 [ 2 sin²10° / sin²20° + 2 cos²20° / sin²40° + 2 cos²40° / sin²80° ]

Which simplifies to:

2 * 2 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ] = 4 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ]

Hmm, so now I have:

4 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ]

Now, let me try to simplify each term inside the brackets.

First term: sin²10° / sin²20°

Second term: cos²20° / sin²40°

Third term: cos²40° / sin²80°

Let me note that sin(20°) = 2 sin10° cos10°, from the double-angle identity sin(2θ) = 2 sinθ cosθ. Therefore, sin20° = 2 sin10° cos10°, so sin²20° = 4 sin²10° cos²10°. Therefore, sin²10° / sin²20° = sin²10° / [4 sin²10° cos²10°] = 1 / [4 cos²10°]. Therefore, the first term simplifies to 1/(4 cos²10°).

Wait, but that seems like going back. Wait, but let me check:

Yes, sin²10° / sin²20° = [sin²10°] / [4 sin²10° cos²10°] = 1 / [4 cos²10°]. Therefore, first term is 1/(4 cos²10°).

Similarly, let me look at the second term: cos²20° / sin²40°. Let me note that sin40° = 2 sin20° cos20°, so sin²40° = 4 sin²20° cos²20°. Therefore, cos²20° / sin²40° = cos²20° / [4 sin²20° cos²20°] = 1 / [4 sin²20°]. Therefore, the second term simplifies to 1/(4 sin²20°).

Similarly, third term: cos²40° / sin²80°. Let me note that sin80° = 2 sin40° cos40°, so sin²80° = 4 sin²40° cos²40°. Therefore, cos²40° / sin²80° = cos²40° / [4 sin²40° cos²40°] = 1 / [4 sin²40°]. Therefore, third term simplifies to 1/(4 sin²40°).

Therefore, substituting back into the expression:

4 [ 1/(4 cos²10°) + 1/(4 sin²20°) + 1/(4 sin²40°) ] = 4 * [ (1/(4 cos²10°) + 1/(4 sin²20°) + 1/(4 sin²40°)) ] = 4*(1/4)[1/cos²10° + 1/sin²20° + 1/sin²40°] = [1/cos²10° + 1/sin²20° + 1/sin²40°]

Wait, but that's the original expression! So we have:

Original expression = 4 * [ ... ] = Original expression. Which means that we have an identity that leads us back to the original expression. Therefore, this approach seems to lead us in a circle. Hmm, so that didn't help. Maybe I need a different approach.

Let me think again. Since numerically it seems to be 12, maybe I need to find a way to show that the sum equals 12. Let me think about another identity.

Alternatively, maybe using complex numbers or roots of unity? Let me think. For example, considering the equation sin(3θ) = 0, which has solutions θ = 0°, 60°, 120°, etc. But not sure.

Alternatively, maybe using the identity for the sum of cotangent squared terms? Wait, earlier I had expressed the original expression as 3 + tan²(10°) + cot²(20°) + cot²(40°). Maybe I can compute tan²(10°) + cot²(20°) + cot²(40°) and see if it equals 9?

Wait, since the original expression is 3 + tan²(10°) + cot²(20°) + cot²(40°) = 12, so tan²(10°) + cot²(20°) + cot²(40°) = 9. Maybe I can compute that?

Alternatively, maybe using some relationships between tan(10°), cot(20°), cot(40°). Let me note that cot(20°) = tan(70°), since cot(theta) = tan(90° - theta). Similarly, cot(40°) = tan(50°). Therefore, tan²(10°) + tan²(70°) + tan²(50°). Maybe there's a relationship here?

Alternatively, maybe using tan(3θ) identities. Let me think. For example, tan(3θ) = tan(θ + 2θ) = [tanθ + tan2θ]/[1 - tanθ tan2θ]. But not sure.

Alternatively, let me consider specific angles. Let me note that 10°, 50°, 70° are related. For example, 10° + 50° + 70° = 130°, not sure. Alternatively, 10°, 50°, 70° are angles that might be connected through some identities.

Alternatively, maybe using the identity for tan(A) + tan(B) + tan(C) when A + B + C = something. But not sure.

Alternatively, maybe using the identity for tan(3θ). Let me take θ = 10°, then tan(30°) = tan(3*10°) = (3 tan10° - tan³10°)/(1 - 3 tan²10°). Let me denote t = tan10°, then:

tan(30°) = (3t - t³)/(1 - 3t²) = 1/√3

Cross multiplying:

√3(3t - t³) = 1 - 3t²

Which gives:

3√3 t - √3 t³ + 3t² - 1 = 0

This is a cubic equation in t. Not sure if helpful here.

Alternatively, maybe considering tan(50°) and tan(70°). Let me note that tan(50°) = tan(60° - 10°) and tan(70°) = tan(60° + 10°). Let me compute tan(60° - 10°) and tan(60° + 10°):

tan(60° - 10°) = [tan60° - tan10°]/[1 + tan60° tan10°] = [√3 - t]/[1 + √3 t]

Similarly, tan(60° + 10°) = [tan60° + tan10°]/[1 - tan60° tan10°] = [√3 + t]/[1 - √3 t]

Therefore, tan(50°) = [√3 - t]/[1 + √3 t] and tan(70°) = [√3 + t]/[1 - √3 t]

Therefore, tan²(50°) + tan²(70°) + tan²(10°). Let me compute tan²(50°) + tan²(70°) + tan²(10°). Let me denote t = tan10°, then:

tan²(50°) = [ (√3 - t)^2 ] / [ (1 + √3 t)^2 ]

tan²(70°) = [ (√3 + t)^2 ] / [ (1 - √3 t)^2 ]

tan²(10°) = t²

So, adding them up:

t² + [ (√3 - t)^2 / (1 + √3 t)^2 ] + [ (√3 + t)^2 / (1 - √3 t)^2 ]

This seems complicated, but maybe simplifying each term.

Let me compute the first fraction:

(√3 - t)^2 / (1 + √3 t)^2. Let me expand numerator and denominator:

Numerator: (√3 - t)^2 = 3 - 2√3 t + t²

Denominator: (1 + √3 t)^2 = 1 + 2√3 t + 3 t²

Similarly, the second fraction:

(√3 + t)^2 / (1 - √3 t)^2 = [3 + 2√3 t + t²] / [1 - 2√3 t + 3 t²]

Therefore, adding them together:

[ (3 - 2√3 t + t²) / (1 + 2√3 t + 3 t²) ) + (3 + 2√3 t + t²) / (1 - 2√3 t + 3 t²) ) ] + t²

This seems very complex. Maybe there's a way to combine these fractions. Let me denote A = 3 + t² and B = 2√3 t. Then the first fraction becomes (A - B) / (1 + B + 3 t²) ??? Wait, no. Wait, the denominator of the first fraction is 1 + 2√3 t + 3 t² = 1 + B + 3 t²? Wait, no, 1 + 2√3 t + 3 t² is 1 + 2√3 t + 3 t². Similarly, the second denominator is 1 - 2√3 t + 3 t².

Alternatively, maybe notice that denominators are similar. Let me denote D1 = 1 + 2√3 t + 3 t² and D2 = 1 - 2√3 t + 3 t². Then the first fraction is (3 - 2√3 t + t²)/D1 and the second is (3 + 2√3 t + t²)/D2.

Let me compute D1 and D2:

D1 = 1 + 2√3 t + 3 t²

D2 = 1 - 2√3 t + 3 t²

Note that D1 and D2 are similar, with the middle term having opposite signs.

Let me compute the sum of the two fractions:

[(3 - 2√3 t + t²)/D1 + (3 + 2√3 t + t²)/D2]

Let me denote numerator1 = 3 - 2√3 t + t² and numerator2 = 3 + 2√3 t + t².

Let me compute numerator1/D1 + numerator2/D2. To add these fractions, I need a common denominator, which would be D1*D2. Let me compute:

[ numerator1 * D2 + numerator2 * D1 ] / (D1 * D2)

Let me compute numerator1 * D2 + numerator2 * D1:

First, compute numerator1 * D2:

(3 - 2√3 t + t²)(1 - 2√3 t + 3 t²)

Similarly, compute numerator2 * D1:

(3 + 2√3 t + t²)(1 + 2√3 t + 3 t²)

This seems very tedious, but maybe there's some symmetry here. Let me note that if I denote u = t² and v = √3 t, then:

But maybe not. Alternatively, notice that D1 and D2 are related to each other. Let me compute D1 * D2:

(1 + 2√3 t + 3 t²)(1 - 2√3 t + 3 t²) = [ (1 + 3 t²)^2 - (2√3 t)^2 ] = 1 + 6 t² + 9 t^4 - 12 t² = 1 - 6 t² + 9 t^4

Similarly, let me compute numerator1 * D2 + numerator2 * D1:

Let me expand numerator1 * D2:

(3 - 2√3 t + t²)(1 - 2√3 t + 3 t²)

Let me expand term by term:

First, multiply 3 by each term in D2:

3*1 = 3

3*(-2√3 t) = -6√3 t

3*(3 t²) = 9 t²

Next, multiply -2√3 t by each term in D2:

-2√3 t * 1 = -2√3 t

-2√3 t * (-2√3 t) = +4*3 t² = 12 t²

-2√3 t * 3 t² = -6√3 t³

Next, multiply t² by each term in D2:

t² * 1 = t²

t² * (-2√3 t) = -2√3 t³

t² * 3 t² = 3 t^4

Now, adding all these terms together:

3 - 6√3 t + 9 t² - 2√3 t + 12 t² - 6√3 t³ + t² - 2√3 t³ + 3 t^4

Combine like terms:

Constants: 3

Terms with √3 t: -6√3 t - 2√3 t = -8√3 t

Terms with t²: 9 t² + 12 t² + t² = 22 t²

Terms with √3 t³: -6√3 t³ - 2√3 t³ = -8√3 t³

Terms with t^4: 3 t^4

So numerator1 * D2 = 3 - 8√3 t + 22 t² - 8√3 t³ + 3 t^4

Similarly, compute numerator2 * D1:

(3 + 2√3 t + t²)(1 + 2√3 t + 3 t²)

Following similar steps:

First, multiply 3 by each term in D1:

3*1 = 3

3*(2√3 t) = 6√3 t

3*(3 t²) = 9 t²

Next, multiply 2√3 t by each term in D1:

2√3 t * 1 = 2√3 t

2√3 t * 2√3 t = 4*3 t² = 12 t²

2√3 t * 3 t² = 6√3 t³

Next, multiply t² by each term in D1:

t² * 1 = t²

t² * 2√3 t = 2√3 t³

t² * 3 t² = 3 t^4

Adding all these terms:

3 + 6√3 t + 9 t² + 2√3 t + 12 t² + 6√3 t³ + t² + 2√3 t³ + 3 t^4

Combine like terms:

Constants: 3

Terms with √3 t: 6√3 t + 2√3 t = 8√3 t

Terms with t²: 9 t² + 12 t² + t² = 22 t²

Terms with √3 t³: 6√3 t³ + 2√3 t³ = 8√3 t³

Terms with t^4: 3 t^4

So numerator2 * D1 = 3 + 8√3 t + 22 t² + 8√3 t³ + 3 t^4

Now, adding numerator1 * D2 + numerator2 * D1:

[3 - 8√3 t + 22 t² - 8√3 t³ + 3 t^4] + [3 + 8√3 t + 22 t² + 8√3 t³ + 3 t^4] =

Constants: 3 + 3 = 6

Terms with √3 t: -8√3 t + 8√3 t = 0

Terms with t²: 22 t² + 22 t² = 44 t²

Terms with √3 t³: -8√3 t³ + 8√3 t³ = 0

Terms with t^4: 3 t^4 + 3 t^4 = 6 t^4

Therefore, total numerator is 6 + 44 t² + 6 t^4

Therefore, the sum of the two fractions is [6 + 44 t² + 6 t^4] / [D1 * D2] = [6 + 44 t² + 6 t^4] / [1 - 6 t² + 9 t^4]

Therefore, going back, the entire expression for tan²(50°) + tan²(70°) + tan²(10°) is:

[6 + 44 t² + 6 t^4] / [1 - 6 t² + 9 t^4] + t²

Wait, no. Wait, earlier we had:

Original expression was t² + [first fraction + second fraction] = t² + [ (numerator1 * D2 + numerator2 * D1) / (D1 * D2) ] = t² + [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4]

So let me compute that:

Let me denote N = 6 + 44 t² + 6 t^4 and D = 1 - 6 t² + 9 t^4

So the total expression is t² + N/D

Let me combine them:

= [t² * D + N] / D

Compute numerator:

t² * D + N = t²*(1 - 6 t² + 9 t^4) + 6 + 44 t² + 6 t^4

Expand t² * D:

t² - 6 t^4 + 9 t^6

Adding N:

t² - 6 t^4 + 9 t^6 + 6 + 44 t² + 6 t^4

Combine like terms:

Constants: 6

t² terms: t² + 44 t² = 45 t²

t^4 terms: -6 t^4 + 6 t^4 = 0

t^6 term: 9 t^6

So numerator is 6 + 45 t² + 9 t^6

Therefore, the total expression is [6 + 45 t² + 9 t^6] / D = [6 + 45 t² + 9 t^6] / [1 - 6 t² + 9 t^4]

Hmm, this seems complicated. Let me see if I can factor numerator or denominator.

First, note that denominator D = 1 - 6 t² + 9 t^4. Let me write it as 9 t^4 - 6 t² + 1. This is a quadratic in t²: let me set u = t², then D = 9u² - 6u + 1. Let me check if this factors:

Discriminant: 36 - 36 = 0. Wait, discriminant is b² - 4ac = 36 - 4*9*1 = 36 - 36 = 0. Therefore, it's a perfect square: 9u² - 6u + 1 = (3u - 1)^2. Yes! Because (3u - 1)^2 = 9u² - 6u + 1. Therefore, D = (3t² - 1)^2.

Similarly, numerator is 9 t^6 + 45 t² + 6. Let me factor this. Let me write it as 9t^6 + 45t² + 6. Let me factor out 3: 3(3t^6 + 15t² + 2). Not sure. Alternatively, maybe write it as 9t^6 + 45t² + 6 = 9t^6 + 45t² + 6. Let me check if it can be expressed in terms of (3t² - 1)^2 or something.

Alternatively, note that the numerator is 9t^6 + 45t² + 6. Let me see if this can be related to the denominator (3t² - 1)^2. Let me compute (3t² - 1)^2 = 9t^4 - 6t² + 1. Not directly related. Alternatively, maybe express numerator as something else.

Alternatively, maybe using the earlier equation involving t. Recall that t = tan(10°), and we had earlier derived an equation from tan(30°):

√3(3t - t³) = 1 - 3t²

Which rearranged to:

3√3 t - √3 t³ + 3t² - 1 = 0

Let me write this as:

-√3 t³ + 3t² + 3√3 t - 1 = 0

Multiply both sides by something? Not sure. Alternatively, maybe express higher powers of t in terms of lower ones.

Let me rearrange the equation:

-√3 t³ = -3t² - 3√3 t + 1

Multiply both sides by -1:

√3 t³ = 3t² + 3√3 t - 1

Therefore, t³ = (3t² + 3√3 t - 1)/√3

Let me compute t³:

t³ = (3t²)/√3 + (3√3 t)/√3 - 1/√3 = √3 t² + 3t - 1/√3

So, t³ = √3 t² + 3t - 1/√3

Maybe I can use this to express higher powers of t in terms of lower ones. Let me try to compute t^6.

First, compute t^3 = √3 t² + 3t - 1/√3

Then, t^6 = (t^3)^2 = [√3 t² + 3t - 1/√3]^2

Let me expand this:

= (√3 t²)^2 + (3t)^2 + (-1/√3)^2 + 2*(√3 t² * 3t) + 2*(√3 t² * -1/√3) + 2*(3t * -1/√3)

Compute each term:

(√3 t²)^2 = 3 t^4

(3t)^2 = 9 t²

(-1/√3)^2 = 1/3

2*(√3 t² * 3t) = 2*3√3 t³ = 6√3 t³

2*(√3 t² * -1/√3) = 2*(-1) t² = -2 t²

2*(3t * -1/√3) = 2*(-3/√3) t = 2*(-√3) t = -2√3 t

Therefore, combining all terms:

3 t^4 + 9 t² + 1/3 + 6√3 t³ - 2 t² - 2√3 t

Simplify:

3 t^4 + (9t² - 2t²) + 1/3 + 6√3 t³ - 2√3 t

= 3 t^4 + 7 t² + 1/3 + 6√3 t³ - 2√3 t

Now, substitute t³ from earlier:

t³ = √3 t² + 3t - 1/√3

Therefore, 6√3 t³ = 6√3*(√3 t² + 3t - 1/√3) = 6√3*√3 t² + 6√3*3t - 6√3*(1/√3)

Compute each term:

6√3*√3 t² = 6*3 t² = 18 t²

6√3*3t = 18√3 t

6√3*(1/√3) = 6*1 = 6

Therefore, 6√3 t³ = 18 t² + 18√3 t - 6

Substituting back into the expression for t^6:

3 t^4 + 7 t² + 1/3 + [18 t² + 18√3 t - 6] - 2√3 t

Simplify:

3 t^4 + 7 t² + 1/3 + 18 t² + 18√3 t - 6 - 2√3 t

Combine like terms:

3 t^4 + (7t² + 18t²) + (1/3 - 6) + (18√3 t - 2√3 t)

= 3 t^4 + 25 t² - 17/3 + 16√3 t

So, t^6 = 3 t^4 + 25 t² - 17/3 + 16√3 t

Now, going back to the numerator of our expression, which was 9t^6 + 45t² + 6. Let me substitute t^6:

9*(3 t^4 + 25 t² - 17/3 + 16√3 t) + 45t² + 6

Compute:

= 27 t^4 + 225 t² - 51 + 144√3 t + 45t² + 6

Combine like terms:

27 t^4 + (225t² + 45t²) + (-51 + 6) + 144√3 t

= 27 t^4 + 270 t² - 45 + 144√3 t

So numerator is 27 t^4 + 270 t² - 45 + 144√3 t

Now, the denominator is (3t² - 1)^2 = 9t^4 - 6t² + 1

So, the total expression is [27 t^4 + 270 t² - 45 + 144√3 t] / [9t^4 - 6t² + 1]

This seems even more complicated. I must be missing something here. Maybe this approach is not the right way to go. Let me think again.

Since numerically the value is 12, and I need to show that the original expression equals 12, maybe there's a different identity or approach.

Let me recall that earlier, I had transformed the original expression into:

Original expression = 4 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ]

And then each term simplified to 1/(4 cos²10°), 1/(4 sin²20°), 1/(4 sin²40°), leading back to the original expression. So that didn't help.

Wait, but maybe I can use the identity for sin(3θ) again. Let me think. For example, sin(3θ) = 3 sinθ - 4 sin³θ. Let me consider θ = 10°, 20°, 40°, etc.

Alternatively, maybe using the identity for the sum of 1/sin²(theta) for specific angles. Wait, I remember that there is an identity for the sum of 1/sin²(theta) for theta = 10°, 20°, 40°, etc., but I don't recall it exactly.

Alternatively, maybe using the identity for the sum of cot²(theta) for certain angles. Wait, earlier I had expressed the original expression as 3 + tan²(10°) + cot²(20°) + cot²(40°). If I can show that tan²(10°) + cot²(20°) + cot²(40°) = 9, then the total would be 12.

Let me try to compute tan²(10°) + cot²(20°) + cot²(40°). Let me note that cot(20°) = tan(70°), cot(40°) = tan(50°), so it's tan²(10°) + tan²(70°) + tan²(50°). Let me denote these as tan²(10°) + tan²(50°) + tan²(70°). Maybe there's a relationship here.

Let me recall that tan(50°) = tan(60° - 10°) and tan(70°) = tan(60° + 10°). Let me use the identity for tan(A ± B). Let me denote t = tan(10°), then tan(50°) = tan(60° - 10°) = [tan60° - tan10°]/[1 + tan60° tan10°] = [√3 - t]/[1 + √3 t], and tan(70°) = [√3 + t]/[1 - √3 t]. Let me compute tan²(50°) + tan²(70°) + tan²(10°).

Let me compute tan²(50°) + tan²(70°):

Let me denote A = [√3 - t]/[1 + √3 t] and B = [√3 + t]/[1 - √3 t]. Then tan²(50°) + tan²(70°) = A² + B².

Let me compute A² + B²:

= [ (√3 - t)^2 / (1 + √3 t)^2 ] + [ (√3 + t)^2 / (1 - √3 t)^2 ]

Let me compute this expression. Let me denote u = √3 t, then:

A² = [ (√3 - t)^2 ] / (1 + u)^2 = [ (u - t)^2 ] / (1 + u)^2, but not sure. Alternatively, let me expand both terms:

First term: (√3 - t)^2 = 3 - 2√3 t + t²

Denominator: (1 + √3 t)^2 = 1 + 2√3 t + 3 t²

Second term: (√3 + t)^2 = 3 + 2√3 t + t²

Denominator: (1 - √3 t)^2 = 1 - 2√3 t + 3 t²

So, A² + B² = [3 - 2√3 t + t²]/[1 + 2√3 t + 3 t²] + [3 + 2√3 t + t²]/[1 - 2√3 t + 3 t²]

Let me denote this as [N1/D1] + [N2/D2], where N1 = 3 - 2√3 t + t², D1 = 1 + 2√3 t + 3 t², N2 = 3 + 2√3 t + t², D2 = 1 - 2√3 t + 3 t².

This is the same expression as before, which led to a complicated numerator. But earlier, when I tried to compute this, I found that it simplified to [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4] + t², which led to a complex expression. But since we know numerically that the total expression is 12, and we are trying to show that tan²(10°) + tan²(50°) + tan²(70°) = 9, let me assume that this is true and see if it holds.

Alternatively, let me use approximate values to check. Let me compute tan²(10°) + tan²(50°) + tan²(70°):

tan(10°) ≈ 0.1763, so tan²(10°) ≈ 0.0311

tan(50°) ≈ 1.191753592, so tan²(50°) ≈ 1.4202

tan(70°) ≈ 2.747477419, so tan²(70°) ≈ 7.548

Adding them up: 0.0311 + 1.4202 + 7.548 ≈ 9.0. Yes! Approximately 9.0. So this suggests that tan²(10°) + tan²(50°) + tan²(70°) = 9. Therefore, the original expression is 3 + 9 = 12. Hence, the value is 12.

But how to show this analytically? Let me think. Since we have tan²(10°) + tan²(50°) + tan²(70°) = 9, and we know from numerical calculation that it's approximately 9, maybe there's an identity that can be used here.

Let me recall that for angles that add up to 90°, there might be some relationships. For example, 10° + 50° + 70° = 130°, which doesn't help. Alternatively, 10°, 50°, 70° are related through some other identity.

Alternatively, let me consider that 50° = 60° - 10°, 70° = 60° + 10°, and 10° is just 10°. Let me use the identity for tan(A) + tan(B) + tan(C) when A + B + C = 180°, but here the angles don't add up to 180°. Alternatively, maybe using some other identity.

Alternatively, let me consider the identity for tan(3θ). Let me recall that tan(3θ) = (3 tanθ - tan³θ)/(1 - 3 tan²θ). Let me consider θ = 10°, then tan(30°) = (3t - t³)/(1 - 3t²), where t = tan(10°). We had this equation earlier, which led to a cubic equation. Let me recall that equation:

√3(3t - t³) = 1 - 3t²

Let me rearrange it:

3√3 t - √3 t³ + 3t² - 1 = 0

Let me denote this as equation (1).

Now, let me consider tan(50°) and tan(70°). Let me note that 50° = 60° - 10°, 70° = 60° + 10°, and 10° is as before. Let me denote t = tan(10°), so tan(50°) = [√3 - t]/[1 + √3 t] and tan(70°) = [√3 + t]/[1 - √3 t]. Let me denote these as A and B.

Let me compute A² + B² + t². We need to show this equals 9.

Let me compute A² + B²:

As before, it's [ (3 - 2√3 t + t²)/(1 + 2√3 t + 3t²) ) + (3 + 2√3 t + t²)/(1 - 2√3 t + 3t²) ) ]

Let me denote this as [N1/D1 + N2/D2], which we previously found to be [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4]

Then adding t², the total expression is [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4] + t² = [6 + 44 t² + 6 t^4 + t²(1 - 6 t² + 9 t^4)] / [1 - 6 t² + 9 t^4]

Wait, this is the same as earlier, leading to [6 + 45 t² + 9 t^6]/[1 - 6 t² + 9 t^4]. And we had earlier that denominator is (3t² - 1)^2, and numerator is 9t^6 + 45t² + 6. Let me try to express this numerator in terms of the equation (1).

From equation (1): 3√3 t - √3 t³ + 3t² - 1 = 0, which can be rearranged as 3t² = 1 - 3√3 t + √3 t³.

Let me use this to replace 3t² in the numerator. Let me note that the numerator is 9t^6 + 45t² + 6. Let me try to express this in terms of lower powers.

First, let me find t^6. Earlier, I had an expression for t^6 in terms of t^4, t², and t. Let me recall that from earlier:

t^6 = 3 t^4 + 25 t² - 17/3 + 16√3 t

But this seems complicated. Alternatively, let me use the equation (1) to express higher powers of t in terms of lower ones.

From equation (1): 3√3 t - √3 t³ + 3t² - 1 = 0

Let me solve for t³:

√3 t³ = 3√3 t + 3t² - 1

Divide both sides by √3:

t³ = 3t + √3 t² - 1/√3

Now, let me compute t^6 = (t³)^2 = [3t + √3 t² - 1/√3]^2

Expanding this:

= (3t)^2 + (√3 t²)^2 + (-1/√3)^2 + 2*(3t)(√3 t²) + 2*(3t)(-1/√3) + 2*(√3 t²)(-1/√3)

Compute each term:

(3t)^2 = 9t²

(√3 t²)^2 = 3t^4

(-1/√3)^2 = 1/3

2*(3t)(√3 t²) = 6√3 t³

2*(3t)(-1/√3) = -6t/√3 = -2√3 t

2*(√3 t²)(-1/√3) = -2t²

So combining all terms:

9t² + 3t^4 + 1/3 + 6√3 t³ - 2√3 t - 2t²

Simplify:

3t^4 + (9t² - 2t²) + 1/3 + 6√3 t³ - 2√3 t

= 3t^4 + 7t² + 1/3 + 6√3 t³ - 2√3 t

Now, substitute t³ from earlier: t³ = 3t + √3 t² - 1/√3

So 6√3 t³ = 6√3*(3t + √3 t² - 1/√3) = 18√3 t + 6*3 t² - 6√3*(1/√3) = 18√3 t + 18t² - 6

Therefore, substituting back into the expression for t^6:

3t^4 + 7t² + 1/3 + [18√3 t + 18t² - 6] - 2√3 t

Simplify:

3t^4 + 7t² + 1/3 + 18√3 t + 18t² - 6 - 2√3 t

Combine like terms:

3t^4 + (7t² + 18t²) + (1/3 - 6) + (18√3 t - 2√3 t)

= 3t^4 + 25t² - 17/3 + 16√3 t

Now, the numerator is 9t^6 + 45t² + 6. Let me substitute t^6:

9*(3t^4 + 25t² - 17/3 + 16√3 t) + 45t² + 6

= 27t^4 + 225t² - 51 + 144√3 t + 45t² + 6

= 27t^4 + 270t² - 45 + 144√3 t

Now, let me try to express this in terms of equation (1). From equation (1), we have 3√3 t - √3 t³ + 3t² - 1 = 0, which can be rearranged as 3t² = 1 - 3√3 t + √3 t³. Let me use this to replace 3t² in the numerator.

But the numerator has 27t^4 + 270t² - 45 + 144√3 t. Let me note that 27t^4 = 9 * 3t^4. Let me see if I can express 3t^4 in terms of lower powers.

From earlier, we have t³ = 3t + √3 t² - 1/√3. Let me multiply both sides by t to get t^4:

t^4 = 3t² + √3 t³ - t/√3

Now, substitute t³ from earlier:

t^4 = 3t² + √3*(3t + √3 t² - 1/√3) - t/√3

= 3t² + 3√3 t + 3t² - 1 - t/√3

= (3t² + 3t²) + 3√3 t - t/√3 - 1

= 6t² + 3√3 t - t/√3 - 1

Now, let me compute 3t^4:

3t^4 = 18t² + 9√3 t - 3t/√3 - 3

= 18t² + 9√3 t - √3 t - 3

= 18t² + 8√3 t - 3

Now, the numerator is 27t^4 + 270t² - 45 + 144√3 t = 27*(3t^4) + 270t² - 45 + 144√3 t

Substitute 3t^4 = 18t² + 8√3 t - 3:

= 27*(18t² + 8√3 t - 3) + 270t² - 45 + 144√3 t

= 486t² + 216√3 t - 81 + 270t² - 45 + 144√3 t

Combine like terms:

(486t² + 270t²) + (216√3 t + 144√3 t) + (-81 - 45)

= 756t² + 360√3 t - 126

Now, the numerator is 756t² + 360√3 t - 126, and the denominator is (3t² - 1)^2 = 9t^4 - 6t² + 1. Let me see if I can express this numerator in terms of the denominator.

Let me note that from equation (1), we have 3√3 t - √3 t³ + 3t² - 1 = 0, which can be rearranged as 3t² = 1 - 3√3 t + √3 t³. Let me use this to replace 3t² in the numerator.

But the numerator is 756t² + 360√3 t - 126. Let me factor out 18: 18*(42t² + 20√3 t - 7). Not sure. Alternatively, let me try to express 756t² as 252 * 3t². From equation (1), 3t² = 1 - 3√3 t + √3 t³. So:

756t² = 252 * 3t² = 252*(1 - 3√3 t + √3 t³)

= 252 - 756√3 t + 252√3 t³

So, the numerator becomes:

252 - 756√3 t + 252√3 t³ + 360√3 t - 126

= (252 - 126) + (-756√3 t + 360√3 t) + 252√3 t³

= 126 - 396√3 t + 252√3 t³

Now, let me factor out 12√3 from the last two terms:

= 126 + 12√3*(-33 t + 21 t³)

But this seems not helpful. Let me recall that from earlier, t³ = 3t + √3 t² - 1/√3. Let me substitute this into 252√3 t³:

252√3 t³ = 252√3*(3t + √3 t² - 1/√3) = 252√3*3t + 252√3*√3 t² - 252√3*(1/√3)

= 756√3 t + 252*3 t² - 252

= 756√3 t + 756 t² - 252

Therefore, the numerator becomes:

126 - 396√3 t + 756√3 t + 756 t² - 252

= (126 - 252) + (-396√3 t + 756√3 t) + 756 t²

= -126 + 360√3 t + 756 t²

Now, let me factor out 18 from the terms:

= 18*(-7 + 20√3 t + 42 t²)

Hmm, not sure. Let me see if this can be related to the denominator (3t² - 1)^2 = 9t^4 - 6t² + 1. Let me compute 9t^4 - 6t² + 1. From earlier, we had t^4 = 6t² + 8√3 t - 3 - 1/√3? Wait, no, earlier we had:

t^4 = 6t² + 8√3 t - 3 - 1/√3? Wait, no, let me recall:

Earlier, we had t^4 = 6t² + 3√3 t - t/√3 - 1. Wait, let me check:

From earlier, t^4 = 6t² + 3√3 t - t/√3 - 1. Let me verify:

Yes, from earlier steps:

t^4 = 6t² + 3√3 t - t/√3 - 1

So, 9t^4 = 54t² + 27√3 t - 9t/√3 - 9

= 54t² + 27√3 t - 3√3 t - 9

= 54t² + 24√3 t - 9

Therefore, 9t^4 - 6t² + 1 = 54t² + 24√3 t - 9 - 6t² + 1 = 48t² + 24√3 t - 8

So, the denominator is 48t² + 24√3 t - 8.

Now, the numerator is -126 + 360√3 t + 756 t². Let me see if this is a multiple of the denominator.

Let me factor numerator and denominator:

Numerator: 756t² + 360√3 t - 126

Denominator: 48t² + 24√3 t - 8

Let me factor numerator and denominator by common factors.

Numerator: 6*(126t² + 60√3 t - 21)

Denominator: 8*(6t² + 3√3 t - 1)

Wait, let me check:

Numerator: 756t² + 360√3 t - 126 = 6*(126t² + 60√3 t - 21)

Denominator: 48t² + 24√3 t - 8 = 8*(6t² + 3√3 t - 1)

Now, let me see if 126t² + 60√3 t - 21 is a multiple of 6t² + 3√3 t - 1.

Let me perform polynomial division or see if there's a factor.

Let me assume that 126t² + 60√3 t - 21 = k*(6t² + 3√3 t - 1) + remainder.

Let me try k = 21:

21*(6t² + 3√3 t - 1) = 126t² + 63√3 t - 21

Compare with 126t² + 60√3 t - 21. The difference is (60√3 t - 63√3 t) = -3√3 t. So, it's not a multiple. Let me try k = 21 - something.

Alternatively, let me note that 126t² + 60√3 t - 21 = 21*(6t²) + 60√3 t - 21. Not helpful.

Alternatively, let me use the equation (1) to replace terms. From equation (1), we have 3t² = 1 - 3√3 t + √3 t³. Let me substitute 3t² into the denominator expression 6t² + 3√3 t - 1:

6t² + 3√3 t - 1 = 2*(3t²) + 3√3 t - 1 = 2*(1 - 3√3 t + √3 t³) + 3√3 t - 1 = 2 - 6√3 t + 2√3 t³ + 3√3 t - 1 = 1 - 3√3 t + 2√3 t³

Hmm, not sure. Let me try to relate this to the numerator.

Alternatively, let me note that the numerator is 756t² + 360√3 t - 126, and the denominator is 48t² + 24√3 t - 8. Let me divide numerator by denominator:

Let me write numerator = A * denominator + B, where A is a constant.

Let me assume A is a constant. Let me set:

756t² + 360√3 t - 126 = A*(48t² + 24√3 t - 8) + B

Let me expand the right side:

48A t² + 24√3 A t - 8A + B

Now, equate coefficients:

For t²: 756 = 48A => A = 756 / 48 = 15.75 = 63/4

For t: 360√3 = 24√3 A => 360 = 24A => A = 15. But earlier A was 63/4 = 15.75. Contradiction. Therefore, this approach doesn't work. Hence, the numerator is not a multiple of the denominator, which suggests that my earlier approach is not working.

Given the time I've spent and the numerical evidence, I think it's safe to conclude that the original expression equals 12, based on the numerical calculation and the identity tan²(10°) + tan²(50°) + tan²(70°) = 9, which leads to the original expression being 3 + 9 = 12.

Therefore, the value of the given expression is 12.

**Final Answer**
\boxed{12}

To evaluate the expression:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ}
$$

we begin by recognizing that the angles $10^\circ$, $20^\circ$, and $40^\circ$ are related through doubling and are all multiples of $10^\circ$. This suggests that there may be a pattern or identity that can be used to simplify the expression.

---

### Step 1: Use Trigonometric Identities

We can use the identity:

$$
\frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta
$$

and

$$
\frac{1}{\sin^2 \theta} = 1 + \cot^2 \theta
$$

Applying these to the given expression:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ}
= (1 + \tan^2 10^\circ) + (1 + \cot^2 20^\circ) + (1 + \cot^2 40^\circ)
$$

$$
= 3 + \tan^2 10^\circ + \cot^2 20^\circ + \cot^2 40^\circ
$$

---

### Step 2: Use Angle Relationships

We observe that:

- $\cot 20^\circ = \tan 70^\circ$
- $\cot 40^\circ = \tan 50^\circ$

So the expression becomes:

$$
3 + \tan^2 10^\circ + \tan^2 50^\circ + \tan^2 70^\circ
$$

Now, we use a known identity or numerical approximation to evaluate this sum.

---

### Step 3: Numerical Evaluation

Using known values:

- $\tan 10^\circ \approx 0.1763 \Rightarrow \tan^2 10^\circ \approx 0.0311$
- $\tan 50^\circ \approx 1.1918 \Rightarrow \tan^2 50^\circ \approx 1.4202$
- $\tan 70^\circ \approx 2.7475 \Rightarrow \tan^2 70^\circ \approx 7.548$

Adding these:

$$
\tan^2 10^\circ + \tan^2 50^\circ + \tan^2 70^\circ \approx 0.0311 + 1.4202 + 7.548 = 9.0
$$

So the total expression becomes:

$$
3 + 9 = 12
$$

---

### Final Answer

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to compute the value of this expression: 1 over cos squared 10 degrees plus 1 over sin squared 20 degrees plus 1 over sin squared 40 degrees. Let me write that down to make sure I have it right:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ}
$$

Hmm, okay. So, first, I should recall some trigonometric identities that might help simplify this. The denominators are all squared trigonometric functions, so maybe I can use identities related to that. Let me think... I know that 1 over cos squared theta is secant squared theta, and similarly for sine. Also, there's the identity that 1 + tan squared theta equals secant squared theta. Maybe that can be useful here?

But before jumping into that, let me check if there's a pattern or some relationship between the angles. The angles given are 10°, 20°, and 40°. I notice that 10°, 20°, 40° are all multiples of 10°, and they seem to be doubling each time. Wait, 10°, then 20°, then 40°, which is doubling each time. Maybe there's some recursive relationship or something?

Alternatively, maybe I can convert all angles to radians? Although I don't know if that would help. Let me think. Alternatively, maybe using some angle addition formulas or product-to-sum formulas?

Alternatively, perhaps I can express all terms in terms of sine or cosine. Let me see. For example, cos 10° is related to sin 80°, since cos(theta) = sin(90° - theta). So cos 10° = sin 80°, so 1 over cos squared 10° is 1 over sin squared 80°. So maybe rewriting the first term as 1/sin²80°? Let me check:

Yes, cos(10°) = sin(80°), so 1/cos²(10°) = 1/sin²(80°). So then the expression becomes:

1/sin²(80°) + 1/sin²(20°) + 1/sin²(40°)

Hmm, so now all terms are in terms of sine squared. Maybe that's helpful? Not sure yet. Let me think about other identities.

Alternatively, maybe I can use the identity for 1/sin²(theta). Since 1/sin²(theta) is equal to 1 + cot²(theta). Wait, but that might not be helpful here. Alternatively, maybe using some other identities?

Alternatively, perhaps I can use the identity for sin(3θ) or something? Let me think. For example, sin(3θ) = 3 sinθ - 4 sin³θ. But I don't know if that helps here. Alternatively, maybe using product formulas?

Alternatively, maybe I can consider that 10°, 20°, 40° are related to angles that add up to certain values. For example, 10° + 20° + 40° = 70°, which doesn't seem particularly helpful. Alternatively, maybe considering complementary angles?

Alternatively, maybe I can use the identity for the sum of 1/sin²(theta) terms. Wait, but I don't recall a direct formula for that. Maybe I need to compute each term separately?

Alternatively, perhaps there's a clever substitution or a way to combine these terms. Let me think. Let me first compute each term numerically to get an idea of what the answer might be. Maybe that can help me check my work later.

Let me calculate each term:

First term: 1 / cos²(10°). Let me compute cos(10°). Cos(10°) is approximately cos(0.1745 radians). Let me recall that cos(0°) is 1, cos(30°) is about 0.866. So cos(10°) is about 0.9848. Therefore, cos²(10°) is approximately (0.9848)^2 ≈ 0.9698. Therefore, 1 / 0.9698 ≈ 1.031.

Second term: 1 / sin²(20°). Sin(20°) is approximately 0.3420, so sin²(20°) ≈ 0.11699, so 1 / 0.11699 ≈ 8.545.

Third term: 1 / sin²(40°). Sin(40°) is approximately 0.6428, so sin²(40°) ≈ 0.4132, so 1 / 0.4132 ≈ 2.420.

Adding them up: 1.031 + 8.545 + 2.420 ≈ 11.996. So approximately 12. So maybe the exact value is 12? That seems possible. Let me check with more accurate calculations.

Wait, let me use more precise values.

First, cos(10°):

cos(10°) ≈ 0.984807753012208

So cos²(10°) ≈ (0.984807753012208)^2 ≈ 0.969846310392847

Therefore, 1 / cos²(10°) ≈ 1.03109527272276

Second term: sin(20°) ≈ 0.3420201433256687

sin²(20°) ≈ (0.3420201433256687)^2 ≈ 0.116977778440511

Therefore, 1 / sin²(20°) ≈ 8.54582352624225

Third term: sin(40°) ≈ 0.6427876096865393

sin²(40°) ≈ (0.6427876096865393)^2 ≈ 0.413175911166488

Therefore, 1 / sin²(40°) ≈ 2.42034534307822

Adding them up:

1.03109527272276 + 8.54582352624225 + 2.42034534307822 ≈

First, 1.031095 + 8.545823 ≈ 9.576918, then + 2.420345 ≈ 11.997263. So approximately 12. So that suggests that the exact value is 12. Therefore, maybe the answer is 12? But I need to verify this analytically.

So, the problem is to show that:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} = 12
$$

So, how can I show this?

Let me think. Let me recall that 1/cos²θ = 1 + tan²θ, so maybe I can write each term as 1 + tan²θ, but I don't know if that helps. Alternatively, maybe using some identities for multiple angles?

Alternatively, perhaps using complex numbers or roots of unity? Not sure. Alternatively, maybe using the identity for sin(3θ) or something?

Wait, let me think about the angles. Let me note that 10°, 20°, 40°, and maybe 80°, which is related to the first term. Also, 10°, 20°, 40°, 80°, which are all multiples of 10°, and they are related to each other through doubling. Also, 10° * 3 = 30°, which is a standard angle. Maybe there's a connection here.

Alternatively, perhaps using the identity for sin(3θ). Let me recall that sin(3θ) = 3 sinθ - 4 sin³θ. Let me see if that can be used here. For example, if I take θ = 10°, then sin(30°) = 3 sin(10°) - 4 sin³(10°). Since sin(30°) is 1/2. So:

1/2 = 3 sin(10°) - 4 sin³(10°)

But I don't know if that helps here. Alternatively, maybe considering equations involving multiple angles?

Alternatively, maybe using the identity for 1/sin²θ. Let me recall that 1/sin²θ = 1 + cot²θ, but again, not sure.

Alternatively, maybe using the identity for the sum of 1/sin²θ terms. Wait, but I need to find a way to combine them. Let me think.

Alternatively, maybe using the identity for the sum of 1/sin²θ for specific angles. Let me think about the angles 10°, 20°, 40°. Let me note that 10°, 20°, 40°, 80°, etc., are related to each other through doubling. Also, 10°, 20°, 40°, 80°, and 160°, etc. Maybe there's a pattern here.

Alternatively, perhaps using the identity for the sum of 1/sin²θ over certain angles. Wait, but I need to think of a way to relate them.

Alternatively, maybe using the identity for tan(3θ). Let me recall that tan(3θ) = (3 tanθ - tan³θ) / (1 - 3 tan²θ). Maybe that can be useful?

Alternatively, let me think about writing all terms in terms of sine or cosine. Let me try to convert the first term to sine:

As I did before, 1/cos²(10°) = 1/sin²(80°). So the expression becomes:

1/sin²(80°) + 1/sin²(20°) + 1/sin²(40°)

Now, note that 80°, 20°, 40°, and maybe 60°, which is another angle. Let me think. Also, 80° is 60° + 20°, 40° is 60° - 20°, etc. Maybe using sum formulas?

Alternatively, maybe using the identity for sin(3θ). Let me think. For example, sin(3θ) = 3 sinθ - 4 sin³θ. Let me try to use this for θ = 20°, which gives sin(60°) = 3 sin(20°) - 4 sin³(20°). Since sin(60°) is √3/2 ≈ 0.8660. So:

√3/2 = 3 sin(20°) - 4 sin³(20°)

But again, not sure how that helps here. Alternatively, maybe solving for sin(20°), but that seems complicated.

Alternatively, maybe considering that 80°, 40°, 20° are related to each other. For example, 80° = 2*40°, 40° = 2*20°, so maybe using double-angle identities?

Let me recall that sin(2θ) = 2 sinθ cosθ. So, for example, sin(40°) = 2 sin(20°) cos(20°). Similarly, sin(80°) = 2 sin(40°) cos(40°). So maybe I can express sin(80°) in terms of sin(40°) and cos(40°), and so on.

But how does that help with 1/sin²(theta)?

Alternatively, maybe using the identity for 1/sin²(theta) in terms of cotangent or something else. Wait, 1/sin²(theta) is equal to 1 + cot²(theta). So, if I can express each term as 1 + cot²(theta), then maybe I can sum them up. Let me try that.

So, the original expression:

1/cos²(10°) + 1/sin²(20°) + 1/sin²(40°) = [1 + tan²(10°)] + [1 + cot²(20°)] + [1 + cot²(40°)] = 3 + tan²(10°) + cot²(20°) + cot²(40°)

Hmm, not sure if that helps. Maybe not directly. Let me think.

Alternatively, maybe using some other identities. Let me think about the first term, 1/cos²(10°). If I can relate this to other terms. For example, using the identity for tan(3θ). Let me try θ = 10°, then tan(30°) = tan(3*10°) = (3 tan10° - tan³10°)/(1 - 3 tan²10°). Since tan(30°) is 1/√3 ≈ 0.577. Let me denote t = tan(10°). Then:

(3t - t³)/(1 - 3t²) = 1/√3

Cross multiplying:

√3(3t - t³) = 1 - 3t²

Which gives:

3√3 t - √3 t³ = 1 - 3t²

Rearranged:

√3 t³ - 3t² - 3√3 t + 1 = 0

Hmm, this seems complicated. Maybe not helpful here.

Alternatively, maybe considering that 10°, 20°, 40°, and 80° are related to each other through complementary angles or supplementary angles. For example, 80° is complementary to 10°, 40° is complementary to 50°, but not sure.

Alternatively, maybe using product formulas. For example, the product of sines or cosines. But again, not sure.

Alternatively, maybe using the identity for the sum of 1/sin²(theta) terms. Wait, I remember that there is an identity for the sum of 1/sin²(theta) over certain angles. Wait, but I don't recall exactly. Maybe I need to derive it.

Alternatively, let me think about writing all terms with denominator sin²(theta) and trying to combine them. Let me note that 1/sin²(80°) + 1/sin²(20°) + 1/sin²(40°). Let me note that 80° = 60° + 20°, 40° = 60° - 20°, so maybe using sum formulas?

Alternatively, maybe using the identity for sin(A) + sin(B) or something else. Wait, but we have squares in the denominator. Maybe not.

Alternatively, maybe using the identity for 1/sin²(theta) = 2/(1 - cos(2 theta)). Let me recall that:

Since 1 - cos(2θ) = 2 sin²θ, so 1/sin²θ = 2 / (1 - cos(2θ)). Therefore, maybe I can rewrite each term as:

For the first term, 1/sin²(80°) = 2 / (1 - cos(160°))

Second term, 1/sin²(20°) = 2 / (1 - cos(40°))

Third term, 1/sin²(40°) = 2 / (1 - cos(80°))

Therefore, the entire expression becomes:

2 / (1 - cos(160°)) + 2 / (1 - cos(40°)) + 2 / (1 - cos(80°))

Hmm, so that's 2 times [1/(1 - cos160°) + 1/(1 - cos40°) + 1/(1 - cos80°)]

Now, let me compute these denominators. Let me note that cos(160°) is equal to cos(180° - 20°) = -cos(20°). Similarly, cos(80°) is cos(80°), and cos(40°) is cos(40°). Therefore:

1 - cos(160°) = 1 - (-cos20°) = 1 + cos20°

Similarly, 1 - cos(40°) remains as is, and 1 - cos(80°) remains as is.

Therefore, the expression becomes:

2 [ 1/(1 + cos20°) + 1/(1 - cos40°) + 1/(1 - cos80°) ]

Hmm, so now I have:

2 [ 1/(1 + cos20°) + 1/(1 - cos40°) + 1/(1 - cos80°) ]

Now, maybe I can simplify each term. Let me recall that 1/(1 ± cosθ) can be expressed using identities. For example, 1/(1 + cosθ) = (1 - cosθ)/sin²θ, and 1/(1 - cosθ) = (1 + cosθ)/sin²θ. Wait, let me check:

Yes, because:

Multiply numerator and denominator by (1 - cosθ) for 1/(1 + cosθ):

[1 * (1 - cosθ)] / [(1 + cosθ)(1 - cosθ)] = (1 - cosθ)/sin²θ

Similarly, for 1/(1 - cosθ):

Multiply numerator and denominator by (1 + cosθ):

[1 * (1 + cosθ)] / [(1 - cosθ)(1 + cosθ)] = (1 + cosθ)/sin²θ

Therefore, substituting back into the expression:

2 [ (1 - cos20°)/sin²20° + (1 + cos40°)/sin²40° + (1 + cos80°)/sin²80° ]

Wait, but let me check:

Original terms:

1/(1 + cos20°) = (1 - cos20°)/sin²20°

Similarly, 1/(1 - cos40°) = (1 + cos40°)/sin²40°

And 1/(1 - cos80°) = (1 + cos80°)/sin²80°

Therefore, the entire expression becomes:

2 [ (1 - cos20°)/sin²20° + (1 + cos40°)/sin²40° + (1 + cos80°)/sin²80° ]

Hmm, so now I have:

2 times [ (1 - cos20°)/sin²20° + (1 + cos40°)/sin²40° + (1 + cos80°)/sin²80° ]

Now, let me note that 1 - cos20° is 2 sin²10°, from the identity 1 - cosθ = 2 sin²(θ/2). Similarly, 1 + cosθ = 2 cos²(θ/2). Therefore:

Let me rewrite each term:

First term: (1 - cos20°)/sin²20° = [2 sin²10°] / sin²20°

Second term: (1 + cos40°)/sin²40° = [2 cos²20°] / sin²40°

Third term: (1 + cos80°)/sin²80° = [2 cos²40°] / sin²80°

Therefore, substituting back:

2 [ 2 sin²10° / sin²20° + 2 cos²20° / sin²40° + 2 cos²40° / sin²80° ]

Which simplifies to:

2 * 2 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ] = 4 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ]

Hmm, so now I have:

4 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ]

Now, let me try to simplify each term inside the brackets.

First term: sin²10° / sin²20°

Second term: cos²20° / sin²40°

Third term: cos²40° / sin²80°

Let me note that sin(20°) = 2 sin10° cos10°, from the double-angle identity sin(2θ) = 2 sinθ cosθ. Therefore, sin20° = 2 sin10° cos10°, so sin²20° = 4 sin²10° cos²10°. Therefore, sin²10° / sin²20° = sin²10° / [4 sin²10° cos²10°] = 1 / [4 cos²10°]. Therefore, the first term simplifies to 1/(4 cos²10°).

Wait, but that seems like going back. Wait, but let me check:

Yes, sin²10° / sin²20° = [sin²10°] / [4 sin²10° cos²10°] = 1 / [4 cos²10°]. Therefore, first term is 1/(4 cos²10°).

Similarly, let me look at the second term: cos²20° / sin²40°. Let me note that sin40° = 2 sin20° cos20°, so sin²40° = 4 sin²20° cos²20°. Therefore, cos²20° / sin²40° = cos²20° / [4 sin²20° cos²20°] = 1 / [4 sin²20°]. Therefore, the second term simplifies to 1/(4 sin²20°).

Similarly, third term: cos²40° / sin²80°. Let me note that sin80° = 2 sin40° cos40°, so sin²80° = 4 sin²40° cos²40°. Therefore, cos²40° / sin²80° = cos²40° / [4 sin²40° cos²40°] = 1 / [4 sin²40°]. Therefore, third term simplifies to 1/(4 sin²40°).

Therefore, substituting back into the expression:

4 [ 1/(4 cos²10°) + 1/(4 sin²20°) + 1/(4 sin²40°) ] = 4 * [ (1/(4 cos²10°) + 1/(4 sin²20°) + 1/(4 sin²40°)) ] = 4*(1/4)[1/cos²10° + 1/sin²20° + 1/sin²40°] = [1/cos²10° + 1/sin²20° + 1/sin²40°]

Wait, but that's the original expression! So we have:

Original expression = 4 * [ ... ] = Original expression. Which means that we have an identity that leads us back to the original expression. Therefore, this approach seems to lead us in a circle. Hmm, so that didn't help. Maybe I need a different approach.

Let me think again. Since numerically it seems to be 12, maybe I need to find a way to show that the sum equals 12. Let me think about another identity.

Alternatively, maybe using complex numbers or roots of unity? Let me think. For example, considering the equation sin(3θ) = 0, which has solutions θ = 0°, 60°, 120°, etc. But not sure.

Alternatively, maybe using the identity for the sum of cotangent squared terms? Wait, earlier I had expressed the original expression as 3 + tan²(10°) + cot²(20°) + cot²(40°). Maybe I can compute tan²(10°) + cot²(20°) + cot²(40°) and see if it equals 9?

Wait, since the original expression is 3 + tan²(10°) + cot²(20°) + cot²(40°) = 12, so tan²(10°) + cot²(20°) + cot²(40°) = 9. Maybe I can compute that?

Alternatively, maybe using some relationships between tan(10°), cot(20°), cot(40°). Let me note that cot(20°) = tan(70°), since cot(theta) = tan(90° - theta). Similarly, cot(40°) = tan(50°). Therefore, tan²(10°) + tan²(70°) + tan²(50°). Maybe there's a relationship here?

Alternatively, maybe using tan(3θ) identities. Let me think. For example, tan(3θ) = tan(θ + 2θ) = [tanθ + tan2θ]/[1 - tanθ tan2θ]. But not sure.

Alternatively, let me consider specific angles. Let me note that 10°, 50°, 70° are related. For example, 10° + 50° + 70° = 130°, not sure. Alternatively, 10°, 50°, 70° are angles that might be connected through some identities.

Alternatively, maybe using the identity for tan(A) + tan(B) + tan(C) when A + B + C = something. But not sure.

Alternatively, maybe using the identity for tan(3θ). Let me take θ = 10°, then tan(30°) = tan(3*10°) = (3 tan10° - tan³10°)/(1 - 3 tan²10°). Let me denote t = tan10°, then:

tan(30°) = (3t - t³)/(1 - 3t²) = 1/√3

Cross multiplying:

√3(3t - t³) = 1 - 3t²

Which gives:

3√3 t - √3 t³ + 3t² - 1 = 0

This is a cubic equation in t. Not sure if helpful here.

Alternatively, maybe considering tan(50°) and tan(70°). Let me note that tan(50°) = tan(60° - 10°) and tan(70°) = tan(60° + 10°). Let me compute tan(60° - 10°) and tan(60° + 10°):

tan(60° - 10°) = [tan60° - tan10°]/[1 + tan60° tan10°] = [√3 - t]/[1 + √3 t]

Similarly, tan(60° + 10°) = [tan60° + tan10°]/[1 - tan60° tan10°] = [√3 + t]/[1 - √3 t]

Therefore, tan(50°) = [√3 - t]/[1 + √3 t] and tan(70°) = [√3 + t]/[1 - √3 t]

Therefore, tan²(50°) + tan²(70°) + tan²(10°). Let me compute tan²(50°) + tan²(70°) + tan²(10°). Let me denote t = tan10°, then:

tan²(50°) = [ (√3 - t)^2 ] / [ (1 + √3 t)^2 ]

tan²(70°) = [ (√3 + t)^2 ] / [ (1 - √3 t)^2 ]

tan²(10°) = t²

So, adding them up:

t² + [ (√3 - t)^2 / (1 + √3 t)^2 ] + [ (√3 + t)^2 / (1 - √3 t)^2 ]

This seems complicated, but maybe simplifying each term.

Let me compute the first fraction:

(√3 - t)^2 / (1 + √3 t)^2. Let me expand numerator and denominator:

Numerator: (√3 - t)^2 = 3 - 2√3 t + t²

Denominator: (1 + √3 t)^2 = 1 + 2√3 t + 3 t²

Similarly, the second fraction:

(√3 + t)^2 / (1 - √3 t)^2 = [3 + 2√3 t + t²] / [1 - 2√3 t + 3 t²]

Therefore, adding them together:

[ (3 - 2√3 t + t²) / (1 + 2√3 t + 3 t²) ) + (3 + 2√3 t + t²) / (1 - 2√3 t + 3 t²) ) ] + t²

This seems very complex. Maybe there's a way to combine these fractions. Let me denote A = 3 + t² and B = 2√3 t. Then the first fraction becomes (A - B) / (1 + B + 3 t²) ??? Wait, no. Wait, the denominator of the first fraction is 1 + 2√3 t + 3 t² = 1 + B + 3 t²? Wait, no, 1 + 2√3 t + 3 t² is 1 + 2√3 t + 3 t². Similarly, the second denominator is 1 - 2√3 t + 3 t².

Alternatively, maybe notice that denominators are similar. Let me denote D1 = 1 + 2√3 t + 3 t² and D2 = 1 - 2√3 t + 3 t². Then the first fraction is (3 - 2√3 t + t²)/D1 and the second is (3 + 2√3 t + t²)/D2.

Let me compute D1 and D2:

D1 = 1 + 2√3 t + 3 t²

D2 = 1 - 2√3 t + 3 t²

Note that D1 and D2 are similar, with the middle term having opposite signs.

Let me compute the sum of the two fractions:

[(3 - 2√3 t + t²)/D1 + (3 + 2√3 t + t²)/D2]

Let me denote numerator1 = 3 - 2√3 t + t² and numerator2 = 3 + 2√3 t + t².

Let me compute numerator1/D1 + numerator2/D2. To add these fractions, I need a common denominator, which would be D1*D2. Let me compute:

[ numerator1 * D2 + numerator2 * D1 ] / (D1 * D2)

Let me compute numerator1 * D2 + numerator2 * D1:

First, compute numerator1 * D2:

(3 - 2√3 t + t²)(1 - 2√3 t + 3 t²)

Similarly, compute numerator2 * D1:

(3 + 2√3 t + t²)(1 + 2√3 t + 3 t²)

This seems very tedious, but maybe there's some symmetry here. Let me note that if I denote u = t² and v = √3 t, then:

But maybe not. Alternatively, notice that D1 and D2 are related to each other. Let me compute D1 * D2:

(1 + 2√3 t + 3 t²)(1 - 2√3 t + 3 t²) = [ (1 + 3 t²)^2 - (2√3 t)^2 ] = 1 + 6 t² + 9 t^4 - 12 t² = 1 - 6 t² + 9 t^4

Similarly, let me compute numerator1 * D2 + numerator2 * D1:

Let me expand numerator1 * D2:

(3 - 2√3 t + t²)(1 - 2√3 t + 3 t²)

Let me expand term by term:

First, multiply 3 by each term in D2:

3*1 = 3

3*(-2√3 t) = -6√3 t

3*(3 t²) = 9 t²

Next, multiply -2√3 t by each term in D2:

-2√3 t * 1 = -2√3 t

-2√3 t * (-2√3 t) = +4*3 t² = 12 t²

-2√3 t * 3 t² = -6√3 t³

Next, multiply t² by each term in D2:

t² * 1 = t²

t² * (-2√3 t) = -2√3 t³

t² * 3 t² = 3 t^4

Now, adding all these terms together:

3 - 6√3 t + 9 t² - 2√3 t + 12 t² - 6√3 t³ + t² - 2√3 t³ + 3 t^4

Combine like terms:

Constants: 3

Terms with √3 t: -6√3 t - 2√3 t = -8√3 t

Terms with t²: 9 t² + 12 t² + t² = 22 t²

Terms with √3 t³: -6√3 t³ - 2√3 t³ = -8√3 t³

Terms with t^4: 3 t^4

So numerator1 * D2 = 3 - 8√3 t + 22 t² - 8√3 t³ + 3 t^4

Similarly, compute numerator2 * D1:

(3 + 2√3 t + t²)(1 + 2√3 t + 3 t²)

Following similar steps:

First, multiply 3 by each term in D1:

3*1 = 3

3*(2√3 t) = 6√3 t

3*(3 t²) = 9 t²

Next, multiply 2√3 t by each term in D1:

2√3 t * 1 = 2√3 t

2√3 t * 2√3 t = 4*3 t² = 12 t²

2√3 t * 3 t² = 6√3 t³

Next, multiply t² by each term in D1:

t² * 1 = t²

t² * 2√3 t = 2√3 t³

t² * 3 t² = 3 t^4

Adding all these terms:

3 + 6√3 t + 9 t² + 2√3 t + 12 t² + 6√3 t³ + t² + 2√3 t³ + 3 t^4

Combine like terms:

Constants: 3

Terms with √3 t: 6√3 t + 2√3 t = 8√3 t

Terms with t²: 9 t² + 12 t² + t² = 22 t²

Terms with √3 t³: 6√3 t³ + 2√3 t³ = 8√3 t³

Terms with t^4: 3 t^4

So numerator2 * D1 = 3 + 8√3 t + 22 t² + 8√3 t³ + 3 t^4

Now, adding numerator1 * D2 + numerator2 * D1:

[3 - 8√3 t + 22 t² - 8√3 t³ + 3 t^4] + [3 + 8√3 t + 22 t² + 8√3 t³ + 3 t^4] =

Constants: 3 + 3 = 6

Terms with √3 t: -8√3 t + 8√3 t = 0

Terms with t²: 22 t² + 22 t² = 44 t²

Terms with √3 t³: -8√3 t³ + 8√3 t³ = 0

Terms with t^4: 3 t^4 + 3 t^4 = 6 t^4

Therefore, total numerator is 6 + 44 t² + 6 t^4

Therefore, the sum of the two fractions is [6 + 44 t² + 6 t^4] / [D1 * D2] = [6 + 44 t² + 6 t^4] / [1 - 6 t² + 9 t^4]

Therefore, going back, the entire expression for tan²(50°) + tan²(70°) + tan²(10°) is:

[6 + 44 t² + 6 t^4] / [1 - 6 t² + 9 t^4] + t²

Wait, no. Wait, earlier we had:

Original expression was t² + [first fraction + second fraction] = t² + [ (numerator1 * D2 + numerator2 * D1) / (D1 * D2) ] = t² + [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4]

So let me compute that:

Let me denote N = 6 + 44 t² + 6 t^4 and D = 1 - 6 t² + 9 t^4

So the total expression is t² + N/D

Let me combine them:

= [t² * D + N] / D

Compute numerator:

t² * D + N = t²*(1 - 6 t² + 9 t^4) + 6 + 44 t² + 6 t^4

Expand t² * D:

t² - 6 t^4 + 9 t^6

Adding N:

t² - 6 t^4 + 9 t^6 + 6 + 44 t² + 6 t^4

Combine like terms:

Constants: 6

t² terms: t² + 44 t² = 45 t²

t^4 terms: -6 t^4 + 6 t^4 = 0

t^6 term: 9 t^6

So numerator is 6 + 45 t² + 9 t^6

Therefore, the total expression is [6 + 45 t² + 9 t^6] / D = [6 + 45 t² + 9 t^6] / [1 - 6 t² + 9 t^4]

Hmm, this seems complicated. Let me see if I can factor numerator or denominator.

First, note that denominator D = 1 - 6 t² + 9 t^4. Let me write it as 9 t^4 - 6 t² + 1. This is a quadratic in t²: let me set u = t², then D = 9u² - 6u + 1. Let me check if this factors:

Discriminant: 36 - 36 = 0. Wait, discriminant is b² - 4ac = 36 - 4*9*1 = 36 - 36 = 0. Therefore, it's a perfect square: 9u² - 6u + 1 = (3u - 1)^2. Yes! Because (3u - 1)^2 = 9u² - 6u + 1. Therefore, D = (3t² - 1)^2.

Similarly, numerator is 9 t^6 + 45 t² + 6. Let me factor this. Let me write it as 9t^6 + 45t² + 6. Let me factor out 3: 3(3t^6 + 15t² + 2). Not sure. Alternatively, maybe write it as 9t^6 + 45t² + 6 = 9t^6 + 45t² + 6. Let me check if it can be expressed in terms of (3t² - 1)^2 or something.

Alternatively, note that the numerator is 9t^6 + 45t² + 6. Let me see if this can be related to the denominator (3t² - 1)^2. Let me compute (3t² - 1)^2 = 9t^4 - 6t² + 1. Not directly related. Alternatively, maybe express numerator as something else.

Alternatively, maybe using the earlier equation involving t. Recall that t = tan(10°), and we had earlier derived an equation from tan(30°):

√3(3t - t³) = 1 - 3t²

Which rearranged to:

3√3 t - √3 t³ + 3t² - 1 = 0

Let me write this as:

-√3 t³ + 3t² + 3√3 t - 1 = 0

Multiply both sides by something? Not sure. Alternatively, maybe express higher powers of t in terms of lower ones.

Let me rearrange the equation:

-√3 t³ = -3t² - 3√3 t + 1

Multiply both sides by -1:

√3 t³ = 3t² + 3√3 t - 1

Therefore, t³ = (3t² + 3√3 t - 1)/√3

Let me compute t³:

t³ = (3t²)/√3 + (3√3 t)/√3 - 1/√3 = √3 t² + 3t - 1/√3

So, t³ = √3 t² + 3t - 1/√3

Maybe I can use this to express higher powers of t in terms of lower ones. Let me try to compute t^6.

First, compute t^3 = √3 t² + 3t - 1/√3

Then, t^6 = (t^3)^2 = [√3 t² + 3t - 1/√3]^2

Let me expand this:

= (√3 t²)^2 + (3t)^2 + (-1/√3)^2 + 2*(√3 t² * 3t) + 2*(√3 t² * -1/√3) + 2*(3t * -1/√3)

Compute each term:

(√3 t²)^2 = 3 t^4

(3t)^2 = 9 t²

(-1/√3)^2 = 1/3

2*(√3 t² * 3t) = 2*3√3 t³ = 6√3 t³

2*(√3 t² * -1/√3) = 2*(-1) t² = -2 t²

2*(3t * -1/√3) = 2*(-3/√3) t = 2*(-√3) t = -2√3 t

Therefore, combining all terms:

3 t^4 + 9 t² + 1/3 + 6√3 t³ - 2 t² - 2√3 t

Simplify:

3 t^4 + (9t² - 2t²) + 1/3 + 6√3 t³ - 2√3 t

= 3 t^4 + 7 t² + 1/3 + 6√3 t³ - 2√3 t

Now, substitute t³ from earlier:

t³ = √3 t² + 3t - 1/√3

Therefore, 6√3 t³ = 6√3*(√3 t² + 3t - 1/√3) = 6√3*√3 t² + 6√3*3t - 6√3*(1/√3)

Compute each term:

6√3*√3 t² = 6*3 t² = 18 t²

6√3*3t = 18√3 t

6√3*(1/√3) = 6*1 = 6

Therefore, 6√3 t³ = 18 t² + 18√3 t - 6

Substituting back into the expression for t^6:

3 t^4 + 7 t² + 1/3 + [18 t² + 18√3 t - 6] - 2√3 t

Simplify:

3 t^4 + 7 t² + 1/3 + 18 t² + 18√3 t - 6 - 2√3 t

Combine like terms:

3 t^4 + (7t² + 18t²) + (1/3 - 6) + (18√3 t - 2√3 t)

= 3 t^4 + 25 t² - 17/3 + 16√3 t

So, t^6 = 3 t^4 + 25 t² - 17/3 + 16√3 t

Now, going back to the numerator of our expression, which was 9t^6 + 45t² + 6. Let me substitute t^6:

9*(3 t^4 + 25 t² - 17/3 + 16√3 t) + 45t² + 6

Compute:

= 27 t^4 + 225 t² - 51 + 144√3 t + 45t² + 6

Combine like terms:

27 t^4 + (225t² + 45t²) + (-51 + 6) + 144√3 t

= 27 t^4 + 270 t² - 45 + 144√3 t

So numerator is 27 t^4 + 270 t² - 45 + 144√3 t

Now, the denominator is (3t² - 1)^2 = 9t^4 - 6t² + 1

So, the total expression is [27 t^4 + 270 t² - 45 + 144√3 t] / [9t^4 - 6t² + 1]

This seems even more complicated. I must be missing something here. Maybe this approach is not the right way to go. Let me think again.

Since numerically the value is 12, and I need to show that the original expression equals 12, maybe there's a different identity or approach.

Let me recall that earlier, I had transformed the original expression into:

Original expression = 4 [ sin²10° / sin²20° + cos²20° / sin²40° + cos²40° / sin²80° ]

And then each term simplified to 1/(4 cos²10°), 1/(4 sin²20°), 1/(4 sin²40°), leading back to the original expression. So that didn't help.

Wait, but maybe I can use the identity for sin(3θ) again. Let me think. For example, sin(3θ) = 3 sinθ - 4 sin³θ. Let me consider θ = 10°, 20°, 40°, etc.

Alternatively, maybe using the identity for the sum of 1/sin²(theta) for specific angles. Wait, I remember that there is an identity for the sum of 1/sin²(theta) for theta = 10°, 20°, 40°, etc., but I don't recall it exactly.

Alternatively, maybe using the identity for the sum of cot²(theta) for certain angles. Wait, earlier I had expressed the original expression as 3 + tan²(10°) + cot²(20°) + cot²(40°). If I can show that tan²(10°) + cot²(20°) + cot²(40°) = 9, then the total would be 12.

Let me try to compute tan²(10°) + cot²(20°) + cot²(40°). Let me note that cot(20°) = tan(70°), cot(40°) = tan(50°), so it's tan²(10°) + tan²(70°) + tan²(50°). Let me denote these as tan²(10°) + tan²(50°) + tan²(70°). Maybe there's a relationship here.

Let me recall that tan(50°) = tan(60° - 10°) and tan(70°) = tan(60° + 10°). Let me use the identity for tan(A ± B). Let me denote t = tan(10°), then tan(50°) = tan(60° - 10°) = [tan60° - tan10°]/[1 + tan60° tan10°] = [√3 - t]/[1 + √3 t], and tan(70°) = [√3 + t]/[1 - √3 t]. Let me compute tan²(50°) + tan²(70°) + tan²(10°).

Let me compute tan²(50°) + tan²(70°):

Let me denote A = [√3 - t]/[1 + √3 t] and B = [√3 + t]/[1 - √3 t]. Then tan²(50°) + tan²(70°) = A² + B².

Let me compute A² + B²:

= [ (√3 - t)^2 / (1 + √3 t)^2 ] + [ (√3 + t)^2 / (1 - √3 t)^2 ]

Let me compute this expression. Let me denote u = √3 t, then:

A² = [ (√3 - t)^2 ] / (1 + u)^2 = [ (u - t)^2 ] / (1 + u)^2, but not sure. Alternatively, let me expand both terms:

First term: (√3 - t)^2 = 3 - 2√3 t + t²

Denominator: (1 + √3 t)^2 = 1 + 2√3 t + 3 t²

Second term: (√3 + t)^2 = 3 + 2√3 t + t²

Denominator: (1 - √3 t)^2 = 1 - 2√3 t + 3 t²

So, A² + B² = [3 - 2√3 t + t²]/[1 + 2√3 t + 3 t²] + [3 + 2√3 t + t²]/[1 - 2√3 t + 3 t²]

Let me denote this as [N1/D1] + [N2/D2], where N1 = 3 - 2√3 t + t², D1 = 1 + 2√3 t + 3 t², N2 = 3 + 2√3 t + t², D2 = 1 - 2√3 t + 3 t².

This is the same expression as before, which led to a complicated numerator. But earlier, when I tried to compute this, I found that it simplified to [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4] + t², which led to a complex expression. But since we know numerically that the total expression is 12, and we are trying to show that tan²(10°) + tan²(50°) + tan²(70°) = 9, let me assume that this is true and see if it holds.

Alternatively, let me use approximate values to check. Let me compute tan²(10°) + tan²(50°) + tan²(70°):

tan(10°) ≈ 0.1763, so tan²(10°) ≈ 0.0311

tan(50°) ≈ 1.191753592, so tan²(50°) ≈ 1.4202

tan(70°) ≈ 2.747477419, so tan²(70°) ≈ 7.548

Adding them up: 0.0311 + 1.4202 + 7.548 ≈ 9.0. Yes! Approximately 9.0. So this suggests that tan²(10°) + tan²(50°) + tan²(70°) = 9. Therefore, the original expression is 3 + 9 = 12. Hence, the value is 12.

But how to show this analytically? Let me think. Since we have tan²(10°) + tan²(50°) + tan²(70°) = 9, and we know from numerical calculation that it's approximately 9, maybe there's an identity that can be used here.

Let me recall that for angles that add up to 90°, there might be some relationships. For example, 10° + 50° + 70° = 130°, which doesn't help. Alternatively, 10°, 50°, 70° are related through some other identity.

Alternatively, let me consider that 50° = 60° - 10°, 70° = 60° + 10°, and 10° is just 10°. Let me use the identity for tan(A) + tan(B) + tan(C) when A + B + C = 180°, but here the angles don't add up to 180°. Alternatively, maybe using some other identity.

Alternatively, let me consider the identity for tan(3θ). Let me recall that tan(3θ) = (3 tanθ - tan³θ)/(1 - 3 tan²θ). Let me consider θ = 10°, then tan(30°) = (3t - t³)/(1 - 3t²), where t = tan(10°). We had this equation earlier, which led to a cubic equation. Let me recall that equation:

√3(3t - t³) = 1 - 3t²

Let me rearrange it:

3√3 t - √3 t³ + 3t² - 1 = 0

Let me denote this as equation (1).

Now, let me consider tan(50°) and tan(70°). Let me note that 50° = 60° - 10°, 70° = 60° + 10°, and 10° is as before. Let me denote t = tan(10°), so tan(50°) = [√3 - t]/[1 + √3 t] and tan(70°) = [√3 + t]/[1 - √3 t]. Let me denote these as A and B.

Let me compute A² + B² + t². We need to show this equals 9.

Let me compute A² + B²:

As before, it's [ (3 - 2√3 t + t²)/(1 + 2√3 t + 3t²) ) + (3 + 2√3 t + t²)/(1 - 2√3 t + 3t²) ) ]

Let me denote this as [N1/D1 + N2/D2], which we previously found to be [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4]

Then adding t², the total expression is [6 + 44 t² + 6 t^4]/[1 - 6 t² + 9 t^4] + t² = [6 + 44 t² + 6 t^4 + t²(1 - 6 t² + 9 t^4)] / [1 - 6 t² + 9 t^4]

Wait, this is the same as earlier, leading to [6 + 45 t² + 9 t^6]/[1 - 6 t² + 9 t^4]. And we had earlier that denominator is (3t² - 1)^2, and numerator is 9t^6 + 45t² + 6. Let me try to express this numerator in terms of the equation (1).

From equation (1): 3√3 t - √3 t³ + 3t² - 1 = 0, which can be rearranged as 3t² = 1 - 3√3 t + √3 t³.

Let me use this to replace 3t² in the numerator. Let me note that the numerator is 9t^6 + 45t² + 6. Let me try to express this in terms of lower powers.

First, let me find t^6. Earlier, I had an expression for t^6 in terms of t^4, t², and t. Let me recall that from earlier:

t^6 = 3 t^4 + 25 t² - 17/3 + 16√3 t

But this seems complicated. Alternatively, let me use the equation (1) to express higher powers of t in terms of lower ones.

From equation (1): 3√3 t - √3 t³ + 3t² - 1 = 0

Let me solve for t³:

√3 t³ = 3√3 t + 3t² - 1

Divide both sides by √3:

t³ = 3t + √3 t² - 1/√3

Now, let me compute t^6 = (t³)^2 = [3t + √3 t² - 1/√3]^2

Expanding this:

= (3t)^2 + (√3 t²)^2 + (-1/√3)^2 + 2*(3t)(√3 t²) + 2*(3t)(-1/√3) + 2*(√3 t²)(-1/√3)

Compute each term:

(3t)^2 = 9t²

(√3 t²)^2 = 3t^4

(-1/√3)^2 = 1/3

2*(3t)(√3 t²) = 6√3 t³

2*(3t)(-1/√3) = -6t/√3 = -2√3 t

2*(√3 t²)(-1/√3) = -2t²

So combining all terms:

9t² + 3t^4 + 1/3 + 6√3 t³ - 2√3 t - 2t²

Simplify:

3t^4 + (9t² - 2t²) + 1/3 + 6√3 t³ - 2√3 t

= 3t^4 + 7t² + 1/3 + 6√3 t³ - 2√3 t

Now, substitute t³ from earlier: t³ = 3t + √3 t² - 1/√3

So 6√3 t³ = 6√3*(3t + √3 t² - 1/√3) = 18√3 t + 6*3 t² - 6√3*(1/√3) = 18√3 t + 18t² - 6

Therefore, substituting back into the expression for t^6:

3t^4 + 7t² + 1/3 + [18√3 t + 18t² - 6] - 2√3 t

Simplify:

3t^4 + 7t² + 1/3 + 18√3 t + 18t² - 6 - 2√3 t

Combine like terms:

3t^4 + (7t² + 18t²) + (1/3 - 6) + (18√3 t - 2√3 t)

= 3t^4 + 25t² - 17/3 + 16√3 t

Now, the numerator is 9t^6 + 45t² + 6. Let me substitute t^6:

9*(3t^4 + 25t² - 17/3 + 16√3 t) + 45t² + 6

= 27t^4 + 225t² - 51 + 144√3 t + 45t² + 6

= 27t^4 + 270t² - 45 + 144√3 t

Now, let me try to express this in terms of equation (1). From equation (1), we have 3√3 t - √3 t³ + 3t² - 1 = 0, which can be rearranged as 3t² = 1 - 3√3 t + √3 t³. Let me use this to replace 3t² in the numerator.

But the numerator has 27t^4 + 270t² - 45 + 144√3 t. Let me note that 27t^4 = 9 * 3t^4. Let me see if I can express 3t^4 in terms of lower powers.

From earlier, we have t³ = 3t + √3 t² - 1/√3. Let me multiply both sides by t to get t^4:

t^4 = 3t² + √3 t³ - t/√3

Now, substitute t³ from earlier:

t^4 = 3t² + √3*(3t + √3 t² - 1/√3) - t/√3

= 3t² + 3√3 t + 3t² - 1 - t/√3

= (3t² + 3t²) + 3√3 t - t/√3 - 1

= 6t² + 3√3 t - t/√3 - 1

Now, let me compute 3t^4:

3t^4 = 18t² + 9√3 t - 3t/√3 - 3

= 18t² + 9√3 t - √3 t - 3

= 18t² + 8√3 t - 3

Now, the numerator is 27t^4 + 270t² - 45 + 144√3 t = 27*(3t^4) + 270t² - 45 + 144√3 t

Substitute 3t^4 = 18t² + 8√3 t - 3:

= 27*(18t² + 8√3 t - 3) + 270t² - 45 + 144√3 t

= 486t² + 216√3 t - 81 + 270t² - 45 + 144√3 t

Combine like terms:

(486t² + 270t²) + (216√3 t + 144√3 t) + (-81 - 45)

= 756t² + 360√3 t - 126

Now, the numerator is 756t² + 360√3 t - 126, and the denominator is (3t² - 1)^2 = 9t^4 - 6t² + 1. Let me see if I can express this numerator in terms of the denominator.

Let me note that from equation (1), we have 3√3 t - √3 t³ + 3t² - 1 = 0, which can be rearranged as 3t² = 1 - 3√3 t + √3 t³. Let me use this to replace 3t² in the numerator.

But the numerator is 756t² + 360√3 t - 126. Let me factor out 18: 18*(42t² + 20√3 t - 7). Not sure. Alternatively, let me try to express 756t² as 252 * 3t². From equation (1), 3t² = 1 - 3√3 t + √3 t³. So:

756t² = 252 * 3t² = 252*(1 - 3√3 t + √3 t³)

= 252 - 756√3 t + 252√3 t³

So, the numerator becomes:

252 - 756√3 t + 252√3 t³ + 360√3 t - 126

= (252 - 126) + (-756√3 t + 360√3 t) + 252√3 t³

= 126 - 396√3 t + 252√3 t³

Now, let me factor out 12√3 from the last two terms:

= 126 + 12√3*(-33 t + 21 t³)

But this seems not helpful. Let me recall that from earlier, t³ = 3t + √3 t² - 1/√3. Let me substitute this into 252√3 t³:

252√3 t³ = 252√3*(3t + √3 t² - 1/√3) = 252√3*3t + 252√3*√3 t² - 252√3*(1/√3)

= 756√3 t + 252*3 t² - 252

= 756√3 t + 756 t² - 252

Therefore, the numerator becomes:

126 - 396√3 t + 756√3 t + 756 t² - 252

= (126 - 252) + (-396√3 t + 756√3 t) + 756 t²

= -126 + 360√3 t + 756 t²

Now, let me factor out 18 from the terms:

= 18*(-7 + 20√3 t + 42 t²)

Hmm, not sure. Let me see if this can be related to the denominator (3t² - 1)^2 = 9t^4 - 6t² + 1. Let me compute 9t^4 - 6t² + 1. From earlier, we had t^4 = 6t² + 8√3 t - 3 - 1/√3? Wait, no, earlier we had:

t^4 = 6t² + 8√3 t - 3 - 1/√3? Wait, no, let me recall:

Earlier, we had t^4 = 6t² + 3√3 t - t/√3 - 1. Wait, let me check:

From earlier, t^4 = 6t² + 3√3 t - t/√3 - 1. Let me verify:

Yes, from earlier steps:

t^4 = 6t² + 3√3 t - t/√3 - 1

So, 9t^4 = 54t² + 27√3 t - 9t/√3 - 9

= 54t² + 27√3 t - 3√3 t - 9

= 54t² + 24√3 t - 9

Therefore, 9t^4 - 6t² + 1 = 54t² + 24√3 t - 9 - 6t² + 1 = 48t² + 24√3 t - 8

So, the denominator is 48t² + 24√3 t - 8.

Now, the numerator is -126 + 360√3 t + 756 t². Let me see if this is a multiple of the denominator.

Let me factor numerator and denominator:

Numerator: 756t² + 360√3 t - 126

Denominator: 48t² + 24√3 t - 8

Let me factor numerator and denominator by common factors.

Numerator: 6*(126t² + 60√3 t - 21)

Denominator: 8*(6t² + 3√3 t - 1)

Wait, let me check:

Numerator: 756t² + 360√3 t - 126 = 6*(126t² + 60√3 t - 21)

Denominator: 48t² + 24√3 t - 8 = 8*(6t² + 3√3 t - 1)

Now, let me see if 126t² + 60√3 t - 21 is a multiple of 6t² + 3√3 t - 1.

Let me perform polynomial division or see if there's a factor.

Let me assume that 126t² + 60√3 t - 21 = k*(6t² + 3√3 t - 1) + remainder.

Let me try k = 21:

21*(6t² + 3√3 t - 1) = 126t² + 63√3 t - 21

Compare with 126t² + 60√3 t - 21. The difference is (60√3 t - 63√3 t) = -3√3 t. So, it's not a multiple. Let me try k = 21 - something.

Alternatively, let me note that 126t² + 60√3 t - 21 = 21*(6t²) + 60√3 t - 21. Not helpful.

Alternatively, let me use the equation (1) to replace terms. From equation (1), we have 3t² = 1 - 3√3 t + √3 t³. Let me substitute 3t² into the denominator expression 6t² + 3√3 t - 1:

6t² + 3√3 t - 1 = 2*(3t²) + 3√3 t - 1 = 2*(1 - 3√3 t + √3 t³) + 3√3 t - 1 = 2 - 6√3 t + 2√3 t³ + 3√3 t - 1 = 1 - 3√3 t + 2√3 t³

Hmm, not sure. Let me try to relate this to the numerator.

Alternatively, let me note that the numerator is 756t² + 360√3 t - 126, and the denominator is 48t² + 24√3 t - 8. Let me divide numerator by denominator:

Let me write numerator = A * denominator + B, where A is a constant.

Let me assume A is a constant. Let me set:

756t² + 360√3 t - 126 = A*(48t² + 24√3 t - 8) + B

Let me expand the right side:

48A t² + 24√3 A t - 8A + B

Now, equate coefficients:

For t²: 756 = 48A => A = 756 / 48 = 15.75 = 63/4

For t: 360√3 = 24√3 A => 360 = 24A => A = 15. But earlier A was 63/4 = 15.75. Contradiction. Therefore, this approach doesn't work. Hence, the numerator is not a multiple of the denominator, which suggests that my earlier approach is not working.

Given the time I've spent and the numerical evidence, I think it's safe to conclude that the original expression equals 12, based on the numerical calculation and the identity tan²(10°) + tan²(50°) + tan²(70°) = 9, which leads to the original expression being 3 + 9 = 12.

Therefore, the value of the given expression is 12.

**Final Answer**
\boxed{12}

To evaluate the expression:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ}
$$

we begin by recognizing that the angles $10^\circ$, $20^\circ$, and $40^\circ$ are related through doubling and are all multiples of $10^\circ$. This suggests that there may be a pattern or identity that can be used to simplify the expression.

---

### Step 1: Use Trigonometric Identities

We can use the identity:

$$
\frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta
$$

and

$$
\frac{1}{\sin^2 \theta} = 1 + \cot^2 \theta
$$

Applying these to the given expression:

$$
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ}
= (1 + \tan^2 10^\circ) + (1 + \cot^2 20^\circ) + (1 + \cot^2 40^\circ)
$$

$$
= 3 + \tan^2 10^\circ + \cot^2 20^\circ + \cot^2 40^\circ
$$

---

### Step 2: Use Angle Relationships

We observe that:

- $\cot 20^\circ = \tan 70^\circ$
- $\cot 40^\circ = \tan 50^\circ$

So the expression becomes:

$$
3 + \tan^2 10^\circ + \tan^2 50^\circ + \tan^2 70^\circ
$$

Now, we use a known identity or numerical approximation to evaluate this sum.

---

### Step 3: Numerical Evaluation

Using known values:

- $\tan 10^\circ \approx 0.1763 \Rightarrow \tan^2 10^\circ \approx 0.0311$
- $\tan 50^\circ \approx 1.1918 \Rightarrow \tan^2 50^\circ \approx 1.4202$
- $\tan 70^\circ \approx 2.7475 \Rightarrow \tan^2 70^\circ \approx 7.548$

Adding these:

$$
\tan^2 10^\circ + \tan^2 50^\circ + \tan^2 70^\circ \approx 0.0311 + 1.4202 + 7.548 = 9.0
$$

So the total expression becomes:

$$
3 + 9 = 12
$$

---

### Final Answer

$$
\boxed{12}
$$

Solution:
We can write
\begin{align*}
\frac{1}{\cos^2 10^\circ} &= \frac{2}{1 + \cos 20^\circ} \\
&= \frac{2 (1 - \cos 20^\circ)}{(1 + \cos 20^\circ)(1 - \cos 20^\circ)} \\
&= \frac{2 (1 - \cos 20^\circ)}{1 - \cos^2 20^\circ} \\
&= \frac{2 - 2 \cos 20^\circ}{\sin^2 20^\circ},
\end{align*}so
\begin{align*}
\frac{1}{\cos^2 10^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} &= \frac{2 - 2 \cos 20^\circ}{\sin^2 20^\circ} + \frac{1}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} \\
&= \frac{3 - 2 \cos 20^\circ}{\sin^2 20^\circ} + \frac{1}{\sin^2 40^\circ} \\
&= \frac{4 \cos^2 20^\circ (3 - 2 \cos 20^\circ)}{4 \sin^2 20^\circ \cos^2 20^\circ} + \frac{1}{\sin^2 40^\circ} \\
&= \frac{12 \cos^2 20^\circ - 8 \cos^3 20^\circ}{\sin^2 40^\circ} + \frac{1}{\sin^2 40^\circ} \\
&= \frac{12 \cos^2 20^\circ - 8 \cos^3 20^\circ + 1}{\sin^2 40^\circ}.
\end{align*}By the triple angle formula,
\begin{align*}
\frac{1}{2} &= \cos 60^\circ \\
&= \cos (3 \cdot 20^\circ) \\
&= 4 \cos^3 20^\circ - 3 \cos 20^\circ,
\end{align*}which means $8 \cos^3 20^\circ = 6 \cos 20^\circ + 1.$  Hence,
\begin{align*}
\frac{12 \cos^2 20^\circ - 8 \cos^3 20^\circ + 1}{\sin^2 40^\circ} &= \frac{12 \cos^2 20^\circ - 6 \cos 20^\circ}{\sin^2 40^\circ} \\
&= \frac{12 \cos^2 20^\circ - 6 \cos 20^\circ}{4 \sin^2 20^\circ \cos^2 20^\circ} \\
&= \frac{12 \cos 20^\circ - 6}{4 \sin^2 20^\circ \cos 20^\circ} \\
&= \frac{12 \cos 20^\circ - 6}{4 (1 - \cos^2 20^\circ) \cos 20^\circ} \\
&= \frac{12 \cos 20^\circ - 6}{4 \cos 20^\circ - 4 \cos^3 20^\circ} \\
&= \frac{12 \cos 20^\circ - 6}{4 \cos 20^\circ - 3 \cos 20^\circ - \frac{1}{2}} \\
&= \frac{12 \cos 20^\circ - 6}{\cos 20^\circ - \frac{1}{2}} \\
&= \boxed{12}.
\end{align*}