Question

Prove the following trignometric identities:

$(se{c}^2\theta -1)(cose{c}^2\theta -1)=1$

Answer 1

We know that,

$se{c}^2\theta -ta{n}^2\theta$ = 1

$cose{c}^2\theta -co{t}^2\theta$ = 1

So,

$(se{c}^2\theta -1)(cose{c}^2\theta -1)$
= $ta{n}^2\theta co{t}^2\theta$

=$(tan\theta cot\theta {)}^2$

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

= 1² = 1