Given $0 \leq x \leq \frac{1}{2}$, then the value of $ an \left(\sin ^{-1}\left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}} ight)-\sin ^{-1} x ight)$ is

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Given $0 \leq x \leq \frac{1}{2}$, then the value of $	an \left(\sin ^{-1}\left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}ight)-\sin ^{-1} xight)$ is

Aicharya · Accepted Answer

$\begin{aligned} & 	an \left[\sin ^{-1}\left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}ight)-\sin ^{-1} xight] \ & =	an \left[\sin ^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}ight)-\sin ^{-1} xight] \ & =	an \left[\sin ^{-1}\left(\frac{\sin 	heta+\cos 	heta}{\sqrt{2}}ight)-	hetaight] \quad \ldots\left[\begin{array}{l}	ext { Put } \sin ^{-1} x=	heta \ \Rightarrow x=\sin 	heta\end{array}ight] \ & =	an \left[\sin ^{-1}\left[\sin \left(	heta+\frac{\pi}{4}ight)ight]-	hetaight] \ & =	an \left(	heta+\frac{\pi}{4}-	hetaight) \ & =	an \frac{\pi}{4}=1\end{aligned}$

Given $0 \leq x \leq \frac{1}{2}$, then the value of $\tan \left(\sin…

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