Given 0 leq x leq frac{1}{2},then the value o | vedclass

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(C) Let the given expression be $E = 	an \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}ight\}-\sin ^{-1} xight]$.Subst

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$ 1 $

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(C) Let the given expression be $E = 	an \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}ight\}-\sin ^{-1} xight]$.Substitute $x = \sin 	heta$,where $	heta = \sin^{-1} x$.Since $0 \leq x \leq \frac{1}{2}$,we have $0 \leq 	heta \leq \frac{\pi}{6}$.The expression inside the $\sin^{-1}$ becomes $\frac{\sin 	heta}{\sqrt{2}} + \frac{\cos 	heta}{\sqrt{2}} = \sin 	heta \cos \frac{\pi}{4} + \cos 	heta \sin \frac{\pi}{4} = \sin \left(	heta + \frac{\pi}{4}ight)$.Now,the expression becomes $E = 	an \left[\sin ^{-1}\left\{\sin \left(	heta + \frac{\pi}{4}ight)ight\} - 	hetaight]$.Since $0 \leq 	heta \leq \frac{\pi}{6}$,then $\frac{\pi}{4} \leq 	heta + \frac{\pi}{4} \leq \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12}$.Since $\frac{\pi}{4} \leq 	heta + \frac{\pi}{4} \leq \frac{5\pi}{12}$,the value lies within the principal range of $\sin^{-1}$,so $\sin^{-1}(\sin(	heta + \frac{\pi}{4})) = 	heta + \frac{\pi}{4}$.Thus,$E = 	an \left(	heta + \frac{\pi}{4} - 	hetaight) = 	an \frac{\pi}{4} = 1$.

Given $0 \leq x \leq \frac{1}{2}$,then the value of $\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$ is:

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