Question

Prove the following identity:
$\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sec \theta -\tan \theta$

Answer 1

LHS$=\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}$
$=\sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \sqrt{\frac{1-\sin \theta }{1+\sin \theta }}}$
$=\sqrt{\frac{{(1-\sin \theta )}^2}{(1-{\sin }^2\theta )}}$
$=\sqrt{\frac{{(1-\sin \theta )}^2}{{\cos }^2\theta }}=\sqrt{\frac{{(1-\sin \theta )}^2}{{\cos }^2\theta }}$
$=\frac{1-\sin \theta }{\cos \theta }$

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

$=\sec \theta -\tan \theta =$ RHS