Question

If $\sqrt{\frac{1-sin\alpha }{1+sin\alpha }}=sec\alpha -tan\alpha$, then the quadants in which $\alpha$ lies are:

• A. 2, 3
• B. 1. 4
• C. 1, 2
• D. 3, 4

Answer 1

The correct option is D 3, 4

$L.H.S=\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }}$

$=\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }}\times \frac{1-\sin \alpha }{1-\sin \alpha }$

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

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

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

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

$=\sec \alpha -\tan \alpha$

$\frac{1-\sin \alpha }{1+\sin \alpha }>0$

$1-\sin \alpha >1+\sin \alpha$

$2\sin \alpha <0$

$\sin \alpha <0$

$\therefore$ alpha lies in quadrants 3,4