Question

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

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

Answer 1

Answer: (B)

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

$cosec\theta =\frac{p+q}{p-q},\sin \theta =\frac{p-q}{p+q}$

$\cot (\frac{\theta }{2}+\frac{\pi }{4})=\frac{\cot \left(\frac{\pi }{4}\right)\cot \left(\frac{\theta }{2}\right)-1}{\cot \left(\frac{\pi }{4}\right)+\cot \left(\frac{\theta }{2}\right)}$

$=\frac{\cot \left(\frac{\theta }{2}\right)-1}{\cot \left(\frac{\theta }{2}\right)+1}$.

$\therefore \sin \theta =\frac{p-q}{p+q}$

$\therefore \cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-(\frac{p-q}{p+q}{)}^2}=\frac{2\sqrt{pq}}{p+q}$

$\cos \theta =2{\cos }^2\frac{\theta }{2}-1$

${\cos }^2\frac{\theta }{2}=\frac{1}{2}\left(\frac{2\sqrt{pq}+p+q}{p+q}\right)=\frac{1}{2}\frac{(\sqrt{p}+\sqrt{q}{)}^2}{p+q}$

${\sin }^2\frac{\theta }{2}=1-\frac{1}{2}\left(\frac{2\sqrt{pq}+p+q}{p+q}\right)=\frac{1}{2}\frac{(\sqrt{p}-\sqrt{q}{)}^2}{p+q}$

${\cot }^2\theta =\frac{(\sqrt{p}+\sqrt{q}{)}^2}{(\sqrt{p}-\sqrt{q}{)}^2}\Rightarrow \cot \theta =\frac{(\sqrt{p}+\sqrt{q})}{(\sqrt{p}-\sqrt{q})}$

$\frac{cot\frac{\theta }{2}-1}{cot\frac{\theta }{2}+1}=\sqrt{\frac{q}{p}}$.