Question

prove
tan^4theta+tan^2theta=sec^4theta-sec^2theta​

Answer 1

Answer:

Use the identity and then try to solve

Step-by-step explanation:

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Answer 2

Answer:

LHS = tan⁴θ + tan²θ

= $\frac{sin^4\theta}{cos^4\theta} +\frac{sin^2\theta}{cos^2\theta}$

= $\frac{sin^4\theta+sin^2\theta cos^2\theta}{cos^4\theta}$

= $\frac{sin^2\theta (sin^2\theta+cos^2\theta)}{cos^4\theta}$

= $\frac{1-cos^2\theta}{cos^4\theta}$

= $\frac{1}{cos^4\theta} -\frac{cos^2\theta}{cos^4\theta}$

= $\frac{1}{cos^4\theta} - \frac{1}{cos^2\theta}$

= sec⁴θ - sec²θ

= RHS

PROVED

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Explanation:

$tan^4\theta = (\frac{sin\theta}{cos\theta} )^4 = \frac{sin^4\theta}{cos^4\theta}$

sin²θ + cos²θ = 1

secθ = $\frac{1}{cos\theta}$