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

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(B) ધારો કે આપેલ પદાવલિ $E = \sin \left[ {{	an }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} ight) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {

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$1$

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(B) ધારો કે આપેલ પદાવલિ $E = \sin \left[ {{	an }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} ight) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} ight) ight]$ છે.$x = 	an 	heta$ લેતા,જેનો અર્થ છે કે $	heta = 	an^{-1} x$.પદાવલિ નીચે મુજબ બનશે:$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]$ત્રિકોણમિતીય નિત્યસમ $\cot 2	heta = \frac{1 - 	an^2 	heta}{2 	an 	heta}$ અને $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$ નો ઉપયોગ કરતા:$E = \sin \left[ {{	an }^{ - 1}}(\cot 2	heta ) + {{\cos }^{ - 1}}(\cos 2	heta ) ight]$કારણ કે $\cot 2	heta = 	an(\frac{\pi}{2} - 2	heta)$:$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$.

$\sin \left\{ {{\tan }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} \right) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \right\}$ ની કિંમત શોધો.

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