Question

Prove that $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sec \theta -\tan \theta$

Answer 1

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

Hence proved.