sin left{ {{tan }^{ - 1}}left( {frac{{1 - {x^ | vedclass

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Question

(B) Let the given expression be $E = \sin \left[ {{	an }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} ight) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}

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$1$

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(B) Let the given expression be $E = \sin \left[ {{	an }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} ight) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight) ight]$.Substitute $x = 	an 	heta$,which implies $	heta = 	an^{-1} x$.The expression becomes:$E = \sin \left[ {{	an }^{ - 1}}\left( {\frac{{1 - {{	an }^2}	heta }}{{2	an 	heta }}} ight) + {{\cos }^{ - 1}}\left( {\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}} ight) ight]$Using trigonometric identities $\cot 2	heta = \frac{1 - 	an^2 	heta}{2 	an 	heta}$ and $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$:$E = \sin \left[ {{	an }^{ - 1}}(\cot 2	heta ) + {{\cos }^{ - 1}}(\cos 2	heta ) ight]$Since $\cot 2	heta = 	an(\frac{\pi}{2} - 2	heta)$:$E = \sin \left[ {{	an }^{ - 1}}\left( 	an \left( \frac{\pi}{2} - 2	heta ight) ight) + {{\cos }^{ - 1}}(\cos 2	heta ) ight]$$E = \sin \left[ \left( \frac{\pi}{2} - 2	heta ight) + 2	heta ight]$$E = \sin \left( \frac{\pi}{2} ight) = 1$.

$\sin \left\{ {{\tan }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} \right) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \right\}$ is equal to

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