sin [cot ^{ - 1}(cos tan ^{ - 1}x)] = | vedclass

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(D) Let $	an^{-1}x = 	heta$,then $	an 	heta = x$. We know that $\cos(	an^{-1}x) = \cos 	heta = \frac{1}{\sqrt{1+x^2}}$. Now,the expression becom

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$\sqrt{\frac{x^2+1}{x^2+2}}$

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(D) Let $	an^{-1}x = 	heta$,then $	an 	heta = x$. We know that $\cos(	an^{-1}x) = \cos 	heta = \frac{1}{\sqrt{1+x^2}}$. Now,the expression becomes $\sin[\cot^{-1}(\frac{1}{\sqrt{1+x^2}})]$. Let $\cot^{-1}(\frac{1}{\sqrt{1+x^2}}) = \phi$,then $\cot \phi = \frac{1}{\sqrt{1+x^2}}$. This implies $	an \phi = \sqrt{1+x^2}$. Using the identity $\sin \phi = \frac{	an \phi}{\sqrt{1+	an^2 \phi}}$,we get: $\sin \phi = \frac{\sqrt{1+x^2}}{\sqrt{1+(1+x^2)}} = \sqrt{\frac{1+x^2}{2+x^2}}$. Thus,the correct option is $D$.

$\sin [\cot ^{ - 1}(\cos \tan ^{ - 1}x)] =$

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