tan left[ {frac{1}{2}{{sin }^{ - 1}}left( {fr | vedclass

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Question

(C) Let $a = 	an 	heta$. Then $	heta = 	an^{-1}(a)$.The expression becomes:$	an \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2	an 	heta }}

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$\frac{{2a}}{{1 - {a^2}}}$

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(C) Let $a = 	an 	heta$. Then $	heta = 	an^{-1}(a)$.The expression becomes:$	an \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }}} ight) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}} ight)} ight]$Using the trigonometric identities $\sin 2	heta = \frac{2	an 	heta}{1 + 	an^2 	heta}$ and $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$:$= 	an \left[ {\frac{1}{2}{{\sin }^{ - 1}}(\sin 2	heta ) + \frac{1}{2}{{\cos }^{ - 1}}(\cos 2	heta )} ight]$Assuming the principal value range for $2	heta$:$= 	an \left[ {\frac{1}{2}(2	heta ) + \frac{1}{2}(2	heta )} ight]$$= 	an (	heta + 	heta) = 	an 2	heta$Using the formula $	an 2	heta = \frac{2	an 	heta}{1 - 	an^2 	heta}$:$= \frac{2a}{1 - a^2}$

$\tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)} \right] = $

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