Question

Prove the following trigonometric identities:

$\frac{tan\theta }{1-cot\theta }+\frac{cot\theta }{1-tan\theta }=1+tan\theta +cot\theta$

Answer 1

Ans:

LHS = tanθ1−1tanθ+cotθ1−tanθtanθ1−1tanθ+cotθ1−tanθ

= tan2θtanθ−1+cotθ1−tanθtan2θtanθ−1+cotθ1−tanθ

= 11−tanθ[1tanθ−tan2θ]11−tanθ[1tanθ−tan2θ]

= 11−tanθ[1−tan3θtanθ]11−tanθ[1−tan3θtanθ]

= 11−tanθ(1−tanθ)(1+tanθ+tan2θ)tanθ11−tanθ(1−tanθ)(1+tanθ+tan2θ)tanθ
[Since , a3−b3=(a−b)(a2+ab+b2)a3−b3=(a−b)(a2+ab+b2)]

= 1+tanθ+tan2θtanθ1+tanθ+tan2θtanθ

= 1tanθ+tanθtanθ+tan2θtanθ1tanθ+tanθtanθ+tan2θtanθ

= 1+tanθ+cotθ1+tanθ+cotθ