Question

If $\cos ec\theta =\frac{p+q}{p-q}$ then $\cot \left(\frac{\pi }{4}+\frac{\theta }{2}\right)=$

• A. $\sqrt{p/q}$
• B. $\sqrt{q/p}$
• C. $\sqrt{pq}$
• D. $pq$

Answer 1

The correct option is B $\sqrt{q/p}$

$Given$
$\cos ec\theta =\frac{p+q}{p-q}$
$formulaused$
$\tan (\frac{\pi }{4}+\theta )=\frac{\tan \left(\frac{\pi }{4}\right)+\tan \theta }{1-\tan \left(\frac{\pi }{4}\right)tan\theta }$
${\cos }^2\theta -{\sin }^2\theta =\cos 2\theta$
$\sin 2\theta =2\sin \theta \cos \theta$
$\cos \theta =\sqrt{1-{\sin }^2\theta }$
$\cos ec\theta =\frac{p+q}{p-q}$
$\Rightarrow \frac{1}{\sin \theta }=\frac{p+q}{p-q}$
$\Rightarrow \sin \theta =\frac{p-q}{p+q}$
$\cos \theta =\sqrt{1-{\left(\frac{p-q}{p+q}\right)}^2}$
$=\frac{\sqrt{{(p+q)}^2-{(p-q)}^2}}{p+q}$
$=\frac{\sqrt{(p^2+q^2+2pq)-(p^2+q^2+2pq)}}{p+q}$
$=\frac{\sqrt{4pq}}{p+q}=\frac{2\sqrt{pq}}{p+q}$
$\cot (\frac{\pi }{4}+\frac{\theta }{2})=\frac{1}{\tan (\frac{\pi }{4}+\frac{\theta }{2})}$
$=\frac{1}{\frac{\tan \left(\frac{\pi }{4}\right)+\tan \left(\frac{\theta }{2}\right)}{1-\tan \left(\frac{\pi }{4}\right)\tan \left(\frac{\theta }{2}\right)}}$
$=\frac{1-\tan \left(\frac{\pi }{4}\right)\tan \left(\frac{\theta }{2}\right)}{\tan \left(\frac{\pi }{4}\right)+\tan \left(\frac{\theta }{2}\right)}$
$=\frac{1-\tan \left(\frac{\theta }{2}\right)}{1+\tan \left(\frac{\theta }{2}\right)}$
$=\frac{1-\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}{1+\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}}$
$=\frac{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}$
$=\frac{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}\times \frac{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}$
$=\frac{{\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}}$
$=\frac{1-\sin \theta }{\cos \theta }$
$=\frac{1-\frac{p-q}{p+q}}{\frac{2\sqrt{pq}}{p+q}}$
$=\frac{2q}{2\sqrt{pq}}=\sqrt{\frac{q}{p}}$