Question

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

Answer 1

Answer:

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Answer 2

Answer:

Given LHS = sec^{4}\theta -sec^{2}\thetasec

4

θ−sec

2

θ

= sec^{2}\theta(sec^{2}\theta-1)sec

2

θ(sec

2

θ−1)

= (1+tan^{2}\theta)(1+tan^{2}\theta-1)(1+tan

2

θ)(1+tan

2

θ−1)

/* By Trigonometric identity:

sec²A = 1+tan²A */

= (1+tan^{2}\theta)\times tan^{2}\theta(1+tan

2

θ)×tan

2

θ

=tan^{2}\theta+tan^{4}\thetatan

2

θ+tan

4

θ

Rearranging the terms, we get

= tan^{4}\theta+tan^{2}\thetatan

4

θ+tan

2

θ

= RHS