Question

Prove that : $\frac{\text{cosec}\theta -1}{\text{cosec}\theta +1}=\frac{{\cos }^2\theta }{{\left(1+\sin \theta \right)}^2}$

Answer 1

Given, $\frac{cosec\theta -1}{cosec\theta +1}$

$=\frac{\frac{1}{sin\theta }-1}{\frac{1}{sin\theta }+1}$

$=\frac{\frac{1-sin\theta }{sin\theta }}{\frac{1+sin\theta }{sin\theta }}$

$=\frac{1-sin\theta }{1+sin\theta }\times \frac{1+sin\theta }{1+sin\theta }$

$=\frac{1-si{n}^2\theta }{(1+sin\theta )}$

$=\frac{co{s}^2\theta }{(1+sin\theta {)}^2}$