Question

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

• A. True
• B. False

Answer 1

The correct option is A True

L.H.S. :

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

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

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

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

$=[\frac{1-\sin A}{\cos A}{]}^2$

$=(\sec A-\tan A{)}^2$ = R.H.S.