MCQ
$\frac{{1 + \sin A - \cos A}}{{1 + \sin A + \cos A}} =$
- A$\sin \frac{A}{2}$
- B$\cos \frac{A}{2}$
- ✓$\tan \frac{A}{2}$
- D$\cot \frac{A}{2}$
$ = \frac{{2\,{{\sin }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\,{{\cos }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}$
$ = \frac{{2\,\,\sin \frac{A}{2}\,\left( {\sin \frac{A}{2} + \cos \frac{A}{2}} \right)}}{{2\,\,\cos \frac{A}{2}\,\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}}$
$= \tan \frac{A}{2}$.
Trick : Put $A = {60^o}.$
$\frac{{1 + (\sqrt 3 /2) - (1/2)}}{{1 + (\sqrt 3 /2) + (1/2)}} = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} = \frac{1}{{\sqrt 3 }}$
which is given by option $(c)$,
$i.e.$ $\tan \frac{{{{60}^o}}}{2} = \frac{1}{{\sqrt 3 }}$
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