The trigonometric identity (cos⁴θ - sin⁴θ)/(1 - tan⁴θ) = cos⁴θ
How to solve the trigonometric identity?
Since (cos⁴θ - sin⁴θ)/(1 - tan⁴θ) = [(cos²θ)² - (sin²θ)²]/[1 - (tan²θ)²]
Using the identity a² - b² = (a + b)(a - b), we have
(cos⁴θ - sin⁴θ)/(1 - tan⁴θ) = [(cos²θ)² - (sin²θ)²]/[1 - (tan²θ)²]
= (cos²θ - sin²θ)(cos²θ + sin²θ)/[(1 - tan²θ)(1 + tan²θ)] =
= (cos²θ - sin²θ) × 1/[(1 - tan²θ)sec²θ] (since (cos²θ + sin²θ) = 1 and 1 + tan²θ = sec²θ)
Also, Using the identity a² - b² = (a + b)(a - b), we have
(cos²θ - sin²θ) × 1/[(1 - tan²θ)sec²θ] = (cosθ - sinθ)(cosθ + sinθ)/[(1 - tanθ)(1 + tanθ)sec²θ]
= (cosθ - sinθ)(cosθ + sinθ)/[(cosθ - sinθ)/cosθ × (cosθ + sinθ)/cosθ × sec²θ]
= (cosθ - sinθ)(cosθ + sinθ)/[(cosθ - sinθ)(cosθ + sinθ)/cos²θ × 1/cos²θ]
= (cosθ - sinθ)(cosθ + sinθ)cos⁴θ/[(cosθ - sinθ)(cosθ + sinθ)]
= 1 × cos⁴θ
= cos⁴θ
So, the trigonometric identity (cos⁴θ - sin⁴θ)/(1 - tan⁴θ) = cos⁴θ
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