Question

If $cosec\theta =\frac{p+q}{(p-q)},thencot(\frac{\pi }{4}+\frac{\theta }{2})$=

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

Answer 1

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

Given, $cosec\theta =\frac{p+q}{p-q}\Rightarrow \frac{1}{sin\theta }=\frac{p+q}{p-q}$
Apply componendo and dividendo
$\frac{1+sin\theta }{1-sin\theta }=\frac{p+q+p-q}{p+q-p+q}$
$\Rightarrow {\left\{\frac{cos\frac{\theta }{2}+sin\frac{\theta }{2}}{cos\frac{\theta }{2}-sin\frac{\theta }{2}}\right\}}^2=\frac{p}{q}\Rightarrow {\left\{\frac{1+tan\frac{\theta }{2}}{1-tan\frac{\theta }{2}}\right\}}^2=\frac{p}{q}$
$\Rightarrow ta{n}^2(\frac{\pi }{4}+\frac{\theta }{2})=\frac{p}{q}\Rightarrow co{s}^2(\frac{\pi }{4}+\frac{\theta }{2})=\frac{q}{p}$
Note: $cot(\frac{\pi }{4}+\frac{\theta }{2})=\sqrt{\frac{q}{p}}only,ifcot(\frac{\pi }{4}+\frac{\theta }{2})$ > 0.