MCQ
If ${\rm{cosec}}\theta = \frac{{p + q}}{{p - q}},$ then $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = $
- A$\sqrt {\frac{p}{q}} $
- ✓$\sqrt {\frac{q}{p}} $
- C$\sqrt {pq} $
- D$pq$
==> $\frac{1}{{\sin \theta }} = \frac{{p + q}}{{p - q}}$
Apply componendo and dividendo
$\frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{{p + q + p - q}}{{p + q - p + q}}$
==> ${\left\{ {\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} - \sin \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$
==> ${\left\{ {\frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$
==> ${\tan ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{p}{q}$
==> ${\cot ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{q}{p}$
Note : $\cot \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \sqrt {\frac{q}{p}} \,{\rm{only,}}\,\,{\rm{if}}$
$\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) > 0$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.