The value of tan left{frac{1}{2} sin ^{-1}lef | vedclass

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Question

(A) Let $x = 	an 	heta$ and $y = 	an \phi$. Substituting these into the expression: $\frac{1}{2} \sin ^{-1} \left( \frac{2 	an 	heta}{1 + 	an^2

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$\frac{x+y}{1-x y}$

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(A) Let $x = 	an 	heta$ and $y = 	an \phi$. Substituting these into the expression: $\frac{1}{2} \sin ^{-1} \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight) = \frac{1}{2} \sin ^{-1} (\sin 2 	heta) = \frac{1}{2} (2 	heta) = 	heta$. Similarly,$\frac{1}{2} \cos ^{-1} \left( \frac{1 - 	an^2 \phi}{1 + 	an^2 \phi} ight) = \frac{1}{2} \cos ^{-1} (\cos 2 \phi) = \frac{1}{2} (2 \phi) = \phi$. Now,the expression becomes $	an (	heta + \phi)$. Using the trigonometric identity $	an (	heta + \phi) = \frac{	an 	heta + 	an \phi}{1 - 	an 	heta 	an \phi}$,we get: $\frac{x + y}{1 - xy}$.

The value of $\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}$ is

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