Question

Prove that:-
$\frac{1-\sin \theta }{1+\sin \theta }=(\sec \theta -\tan \theta {)}^2$

Answer 1

We have,

$R.H.S.$

$(\sec \theta -\tan \theta {)}^2$

$\Rightarrow {(\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta })}^2$

$\Rightarrow {\left(\frac{1-\sin \theta }{\cos \theta }\right)}^2$

$\Rightarrow \frac{{(1-\sin \theta )}^2}{{\cos }^2\theta }$

$\Rightarrow \frac{{(1-\sin \theta )}^2}{1-{\sin }^2\theta }$

$\Rightarrow \frac{(1-\sin \theta )(1-\sin \theta )}{(1+\sin \theta )(1-\sin \theta )}$

$\Rightarrow \frac{(1-\sin \theta )}{(1+\sin \theta )}$

L.H.S.

Hence, proved.