If cos ec,theta = frac{{p + q}}{{p | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\cos ec\,	heta  = \frac{{p + q}}{{p - q}},$ $\sin 	heta  = \frac{{p - q}}{{p + q}}$

$\cos 	heta$

$=\pm \sqrt{1-\sin ^{2} 	he

Vedclass · Accepted Answer

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

Vedclass · Answer

$\cos ec\,	heta  = \frac{{p + q}}{{p - q}},$ $\sin 	heta  = \frac{{p - q}}{{p + q}}$

$\cos 	heta$

$=\pm \sqrt{1-\sin ^{2} 	heta}=\sqrt{1-\left(\frac{p-q}{p+q}ight)^{2}}=\frac{2 \sqrt{p q}}{(p+q)}$

$\left|\cot \left(\frac{\pi}{4}+\frac{	heta}{2}ight)ight|=\frac{\cot \frac{\pi}{4} \cot \frac{	heta}{2}-1}{\cot \frac{\pi}{4}+\cot \frac{	heta}{2}}$

$=\frac{\cot \frac{	heta}{2}-1}{\cot \frac{	heta}{2}+1}=\frac{\cos \frac{	heta}{2}-\sin \frac{	heta}{2}}{\cos \frac{	heta}{2}+\sin \frac{	heta}{2}}$

On rationalizing denominator, we get

${\left( {\frac{{\cos \frac{	heta }{2} - \sin \frac{	heta }{2}}}{{\cos \frac{	heta }{2} + \sin \frac{	heta }{2}}}} ight)}$  ${\left( {\frac{{\cos \frac{	heta }{2} + \sin \frac{	heta }{2}}}{{\cos \frac{	heta }{2} + \sin \frac{	heta }{2}}}} ight)}$

$ = \,\frac{{\cos \,	heta }}{{{{\sin }^2}\frac{	heta }{2} + {{\cos }^2}\frac{	heta }{2} + 2\sin \frac{	heta }{2}\cos \frac{	heta }{2}}}$

$ = \left| {\frac{{\cos 	heta }}{{1 + \sin 	heta }}} ight|\, = \,\frac{{2\sqrt {pq} /(p + q)}}{{1 + \,\frac{{(p - q)}}{{(p + q)}}}}$

$=\frac{\sqrt{p q}}{p}=\sqrt{\frac{q}{p}}$

If $\cos ec\,\theta = \frac{{p + q}}{{p - q}}$ $\left( {p \ne q \ne 0} \right)$, then $\left| {\cot \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right)} \right|$ is equal to

Similar Questions

Number of solutions of $8cosx$ = $x$ will be

If $\theta $ and $\phi $ are acute satisfying $\sin \theta = \frac{1}{2},$ $\cos \phi = \frac{1}{3},$ then $\theta + \phi \in $

The most general value of $\theta $ which will satisfy both the equations $\sin \theta = - \frac{1}{2}$ and $\tan \theta = \frac{1}{{\sqrt 3 }}$ is

Number of solutions of the equation $2^x + x = 2^{sin \ x} + \sin x$ in $[0,10\pi ]$ is -

If $\left| {\,\begin{array}{*{20}{c}}{\cos (A + B)}&{ - \sin (A + B)}&{\cos 2B}\\{\sin A}&{\cos A}&{\sin B}\\{ - \cos A}&{\sin A}&{\cos B}\end{array}\,} \right| = 0$, then $B =$