1 - A 1 + A = - Vidyadip

MCQ

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

A
$\sec A + \tan A$
B
$\tan \left( {\frac{\pi }{4} - A} \right)$
C
$\tan \left( {\frac{\pi }{4} + \frac{A}{2}} \right)$
✓
$\tan \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$

✓

Answer

Correct option: D.

$\tan \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$

d
(d) $\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = \sqrt {\frac{{1 - \cos \left( {\frac{\pi }{2} - A} \right)}}{{1 + \cos \left( {\frac{\pi }{2} - A} \right)}}} $

$ = \sqrt {\frac{{2{{\sin }^2}\left( {\frac{\pi }{4} - \frac{A}{2}} \right)}}{{2{{\cos }^2}\left( {\frac{\pi }{4} - \frac{A}{2}} \right)}}} = \tan \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$.

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