If operatorname{cosec} theta = frac{p+q}{p-q} | vedclass

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(A) Given $\operatorname{cosec} 	heta = \frac{p+q}{p-q}$.We know $\sin 	heta = \frac{p-q}{p+q}$.Using the formula $\sin 	heta = \frac{2 	an(	heta

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

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(A) Given $\operatorname{cosec} 	heta = \frac{p+q}{p-q}$.We know $\sin 	heta = \frac{p-q}{p+q}$.Using the formula $\sin 	heta = \frac{2 	an(	heta/2)}{1 + 	an^2(	heta/2)}$,we have $\frac{2 	an(	heta/2)}{1 + 	an^2(	heta/2)} = \frac{p-q}{p+q}$.Inverting both sides: $\frac{1 + 	an^2(	heta/2)}{2 	an(	heta/2)} = \frac{p+q}{p-q}$.Applying Componendo and Dividendo:$\frac{1 + 	an^2(	heta/2) + 2 	an(	heta/2)}{1 + 	an^2(	heta/2) - 2 	an(	heta/2)} = \frac{(p+q) + (p-q)}{(p+q) - (p-q)}$.$\frac{(1 + 	an(	heta/2))^2}{(1 - 	an(	heta/2))^2} = \frac{2p}{2q} = \frac{p}{q}$.Taking the square root: $\frac{1 + 	an(	heta/2)}{1 - 	an(	heta/2)} = \sqrt{\frac{p}{q}}$.Since $	an(\pi/4) = 1$,this is $	an(\pi/4 + 	heta/2) = \sqrt{\frac{p}{q}}$.Therefore,$\cot(\pi/4 + 	heta/2) = \frac{1}{	an(\pi/4 + 	heta/2)} = \sqrt{\frac{q}{p}}$.

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

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