If text{cosec} theta = frac{p + q}{p - q},the | vedclass

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(B) Given,$	ext{cosec} 	heta = \frac{p + q}{p - q}$.$\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying componendo and dividendo:$\frac{1 + \sin

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

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(B) Given,$	ext{cosec} 	heta = \frac{p + q}{p - q}$.$\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying componendo and dividendo:$\frac{1 + \sin 	heta}{1 - \sin 	heta} = \frac{(p + q) + (p - q)}{(p + q) - (p - q)} = \frac{2p}{2q} = \frac{p}{q}$.We know that $1 + \sin 	heta = \left( \cos \frac{	heta}{2} + \sin \frac{	heta}{2} ight)^2$ and $1 - \sin 	heta = \left( \cos \frac{	heta}{2} - \sin \frac{	heta}{2} ight)^2$.So,$\left( \frac{\cos \frac{	heta}{2} + \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} - \sin \frac{	heta}{2}} ight)^2 = \frac{p}{q}$.Dividing numerator and denominator by $\cos \frac{	heta}{2}$,we get:$\left( \frac{1 + 	an \frac{	heta}{2}}{1 - 	an \frac{	heta}{2}} ight)^2 = \frac{p}{q}$.Using the formula $	an \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \frac{1 + 	an \frac{	heta}{2}}{1 - 	an \frac{	heta}{2}}$,we have:$	an^2 \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \frac{p}{q}$.Taking the reciprocal:$\cot^2 \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \frac{q}{p}$.Therefore,$\cot \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \sqrt{\frac{q}{p}}$.

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

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