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यदि ${\rm{cosec}}\theta = \frac{{p + q}}{{p - q}},$ तब $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = $
$\sqrt {\frac{p}{q}} $
$\sqrt {\frac{q}{p}} $
$\sqrt {pq} $
$pq$
Solution
(b) दिया है, ${\rm{cosec}}\theta = \frac{{p + q}}{{p – q}}$
==> $\frac{1}{{\sin \theta }} = \frac{{p + q}}{{p – q}}$
योगान्तरानुपात नियम से,
$\frac{{1 + \sin \theta }}{{1 – \sin \theta }} = \frac{{p + q + p – q}}{{p + q – p + q}}$
==> ${\left\{ {\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} – \sin \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$
==> ${\left\{ {\frac{{1 + \tan \frac{\theta }{2}}}{{1 – \tan \frac{\theta }{2}}}} \right\}^2} = \frac{p}{q}$
==> ${\tan ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{p}{q}$
==> ${\cot ^2}\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \frac{q}{p}$
नोट : $\cot \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = \sqrt {\frac{q}{p}}$ सिर्फ, यदि $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) > 0$.
