Question

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

Answer 1

Consider L.H.S

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

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

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

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

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

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

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

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

$=R.H.S$

LHS=RHS

Hence proved.