If {sin ^{ - 1}}left( {frac{{2a}}{{1 + {a^2}} | vedclass

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(D) Given the equation: ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} ight) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} ight) = 2{	an ^{

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$\frac{{a + b}}{{1 - ab}}$

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(D) Given the equation: ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} ight) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} ight) = 2{	an ^{ - 1}}x$Substitute $a = 	an 	heta$ and $b = 	an \phi$:${\sin ^{ - 1}}\left( {\frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }}} ight) + {\sin ^{ - 1}}\left( {\frac{{2	an \phi }}{{1 + {{	an }^2}\phi }}} ight) = 2{	an ^{ - 1}}x$Using the identity $\sin 2	heta = \frac{2	an 	heta}{1 + 	an^2 	heta}$,we get:${\sin ^{ - 1}}(\sin 2	heta) + {\sin ^{ - 1}}(\sin 2\phi) = 2{	an ^{ - 1}}x$Simplifying the expression:$2	heta + 2\phi = 2{	an ^{ - 1}}x$Dividing by $2$:$	heta + \phi = {	an ^{ - 1}}x$Taking the tangent of both sides:$x = 	an(	heta + \phi)$Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$:$x = \frac{	an 	heta + 	an \phi}{1 - 	an 	heta 	an \phi}$Substituting $a$ and $b$ back:$x = \frac{a + b}{1 - ab}$

If ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x,$ then $x = $

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