यदि {rm{cosec}}theta = frac{{p + q}}{{p - q}} | vedclass

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Question

(b) दिया है,  ${m{cosec}}	heta = \frac{{p + q}}{{p - q}}$

==> $\frac{1}{{\sin 	heta }} = \frac{{p + q}}{{p - q}}$

योगान्तर

Vedclass · Accepted Answer

$\sqrt {\frac{q}{p}} $

Vedclass · Answer

(b) दिया है,  ${m{cosec}}	heta = \frac{{p + q}}{{p - q}}$

==> $\frac{1}{{\sin 	heta }} = \frac{{p + q}}{{p - q}}$

योगान्तरानुपात नियम से, 

$\frac{{1 + \sin 	heta }}{{1 - \sin 	heta }} = \frac{{p + q + p - q}}{{p + q - p + q}}$

==> ${\left\{ {\frac{{\cos \frac{	heta }{2} + \sin \frac{	heta }{2}}}{{\cos \frac{	heta }{2} - \sin \frac{	heta }{2}}}} ight\}^2} = \frac{p}{q}$ 

==> ${\left\{ {\frac{{1 + 	an \frac{	heta }{2}}}{{1 - 	an \frac{	heta }{2}}}} ight\}^2} = \frac{p}{q}$

==> ${	an ^2}\left( {\frac{\pi }{4} + \frac{	heta }{2}} ight) = \frac{p}{q}$ 

==> ${\cot ^2}\left( {\frac{\pi }{4} + \frac{	heta }{2}} ight) = \frac{q}{p}$

नोट : $\cot \left( {\frac{\pi }{4} + \frac{	heta }{2}} ight) = \sqrt {\frac{q}{p}}$ सिर्फ, यदि  $\cot \,\left( {\frac{\pi }{4} + \frac{	heta }{2}} ight) > 0$.

यदि ${\rm{cosec}}\theta = \frac{{p + q}}{{p - q}},$ तब $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = $

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