If cosec θ=p+qp−q then cot π4+θ2=

cosecθ=p+qp−q⇒1sinθ=p+qp−q⇒1+sinθ1−sinθ=pq⇒cos2θ/2+sin2θ/2+2cosθ/2sinθ/2cos2θ/2+sin2θ/2−2cosθ/2sinθ/2=pq⇒cosθ/2+sinθ/2cosθ/2−sinθ/2=pq⇒tanπ4+θ2=pq⇒cotπ4+θ2=qp

If cosec θ=p+qp−q then cot π4+θ2=

cosecθ=p+qp−q⇒1sinθ=p+qp−q⇒1+sinθ1−sinθ=pq⇒cos2θ/2+sin2θ/2+2cosθ/2sinθ/2cos2θ/2+sin2θ/2−2cosθ/2sinθ/2=pq⇒cosθ/2+sinθ/2cosθ/2−sinθ/2=pq⇒tanπ4+θ2=pq⇒cotπ4+θ2=qp

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$I f \cos e c θ = \frac{p + q}{p - q} t h e n \cot (\frac{π}{4} + \frac{θ}{2}) =$

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$\sqrt{p q}$

$\sqrt{q / p}$

$\sqrt{p / q}$

$p q$

answer is B.

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Detailed Solution

$\begin{array}{l} \cos e c θ = \frac{p + q}{p - q} \Rightarrow \frac{1}{\sin θ} = \frac{p + q}{p - q} \Rightarrow \frac{1 + \sin θ}{1 - \sin θ} = \frac{p}{q} \Rightarrow \frac{\cos^{2} θ / 2 + \sin^{2} θ / 2 + 2 \cos θ / 2 \sin θ / 2}{\cos^{2} θ / 2 + \sin^{2} θ / 2 - 2 \cos θ / 2 \sin θ / 2} = \frac{p}{q} \\ \Rightarrow \frac{\cos θ / 2 + \sin θ / 2}{\cos θ / 2 - \sin θ / 2} = \sqrt{[\frac{p}{q}]} \Rightarrow \tan (\frac{π}{4} + \frac{θ}{2}) = \sqrt{[\frac{p}{q}]} \Rightarrow \cot (\frac{π}{4} + \frac{θ}{2}) = \sqrt{[\frac{q}{p}]} \end{array}$