Question

2sec^20 - sec^40 - 2cosec^20 + cosec^40 = cot^40 - tan^40

'0' is theta .

Pls help me out with this

Answer 1

From the RHS:

Cot^4θ-tan^4θ

=(cos^2θ/sin^2θ)*(cos^2θ/sin^2θ)-(sec^2-1)*(sec^2-1)

=(1-sin^2θ)/(sin^2(θ))*(1-sin^2θ)/(sin^2(θ))-(sec^4θ+2sec^2θ+1)

=(cosec^2θ-1)*(cosec^2θ-1)-sec^4θ+2sec^2θ-1

=cosec^4θ-2cosec^2θ+1-sec^4θ+2sec^2θ-1

=cosec^4θ-2cosec^2θ-sec^4θ+2sec^2θ

=LHS

Explanation:

Answer 2

Answer: see proof below

Explanation:

Use the following Pythagorean Identities:

sec²A = tan²A + 1

csc²A = 1 + cot²A

Use group factoring: a(x + y) + b(x + y) = (a + b)(x + y)

Proof LHS → RHS

Given: 2sec²Ф - sec⁴Ф - 2csc²Ф + csc⁴Ф

Pythagorean: 2(tan²Ф + 1) - (tan²Ф + 1)² + -2(1 + cot²Ф) + (1 + cot²Ф)²

Group Factoring: (tan²Ф + 1)(2 - tan²Ф - 1) + (1 + cot²Ф)(-2 + 1 + cot²Ф)

Simplified: (tan²Ф + 1)(1 - tan²Ф) + (1 + cot²Ф)(cot²Ф - 1)

Multiplied: tan²Ф - tan⁴Ф + 1 - tan²Ф + cot²Ф - 1 + cot⁴Ф - cot²Ф

Simplified: 1 - tan⁴Ф + cot⁴Ф - 1

= - tan⁴Ф + cot⁴Ф

= cot⁴Ф - tan⁴Ф

cot⁴Ф - tan⁴Ф = cot⁴Ф - tan⁴Ф
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