Question

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

Answer 1

R.H.S

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

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

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

We know that

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

Therefore,

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

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

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

L.H.S

Hence, proved.