If csc theta = frac{p + q}{p - q},then cot le | vedclass

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(B) Given,$\csc 	heta = \frac{p + q}{p - q}$.This implies $\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying the componendo and dividendo rule:$\

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

Vedclass · Answer

(B) Given,$\csc 	heta = \frac{p + q}{p - q}$.This implies $\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying the componendo and dividendo rule:$\frac{1 + \sin 	heta}{1 - \sin 	heta} = \frac{(p + q) + (p - q)}{(p + q) - (p - q)} = \frac{2p}{2q} = \frac{p}{q}$.We know that $1 + \sin 	heta = (\cos \frac{	heta}{2} + \sin \frac{	heta}{2})^2$ and $1 - \sin 	heta = (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})^2$.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 the 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(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,where $A = \frac{\pi}{4}$ and $B = \frac{	heta}{2}$:$	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 $\csc \theta = \frac{p + q}{p - q}$,then $\cot \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = $

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