Question

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

Answer 1

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

$=\sqrt{\frac{\left(1-\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}}$

$=\sqrt{\frac{{\left(1-\sin \theta \right)}^2}{1-{\sin }^2\theta }}$

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

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

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

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