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

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(A) Let $\sin^{-1} x = 	heta$,which implies $x = \sin 	heta$. Since $0 \leq x \leq \frac{1}{2}$,we have $0 \leq 	heta \leq \frac{\pi}{6}$.Then $\sq

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$1$

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(A) Let $\sin^{-1} x = 	heta$,which implies $x = \sin 	heta$. Since $0 \leq x \leq \frac{1}{2}$,we have $0 \leq 	heta \leq \frac{\pi}{6}$.Then $\sqrt{1-x^2} = \cos 	heta$.The expression becomes $	an \left[ \sin^{-1} \left( \frac{\sin 	heta}{\sqrt{2}} + \frac{\cos 	heta}{\sqrt{2}} ight) - 	heta ight]$.Using the identity $\sin 	heta + \cos 	heta = \sqrt{2} \sin \left( 	heta + \frac{\pi}{4} ight)$,we get:$\frac{\sin 	heta + \cos 	heta}{\sqrt{2}} = \sin \left( 	heta + \frac{\pi}{4} ight)$.Substituting this into the expression:$= 	an \left[ \sin^{-1} \left( \sin \left( 	heta + \frac{\pi}{4} ight) ight) - 	heta ight]$.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 this range is within $[-\frac{\pi}{2}, \frac{\pi}{2}]$,$\sin^{-1}(\sin(	heta + \frac{\pi}{4})) = 	heta + \frac{\pi}{4}$.Thus,the expression simplifies to $	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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