Answer:
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
Step-by-step explanation:
Answer: see proof below
Step-by-step 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 [tex]\checkmark[/tex]
