MCQ
$\frac{{\cos A}}{{1 - \sin A}} = $
- A$\sec A - \tan A$
- B${\rm{cosec}}\,A + \cot A$
- 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) $\frac{{\cos A}}{{1 - \sin A}} = \frac{{\cos A(1 + \sin A)}}{{{{\cos }^2}A}} = \frac{{(1 + \sin A)}}{{\cos A}}$
(d) $\frac{{\cos A}}{{1 - \sin A}} = \frac{{\cos A(1 + \sin A)}}{{{{\cos }^2}A}} = \frac{{(1 + \sin A)}}{{\cos A}}$
$ = \frac{{{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}^2}}}{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)\,\left( {\cos \frac{A}{2} - \sin \frac{A}{2}} \right)}} $
$= \frac{{\cos \frac{A}{2} + \sin \frac{A}{2}}}{{\cos \frac{A}{2} - \sin \frac{A}{2}}}$
$ = \frac{{1 + \tan \frac{A}{2}}}{{1 - \tan \frac{A}{2}}}$, $\left( {{\rm{Dividing}}\,{N^r}\,{\rm{and}}\,{D^r}\,{\rm{by}}\,\cos \frac{A}{2}} \right)$
$ = \tan \left( {\frac{\pi }{4} + \frac{A}{2}} \right)$.
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