Solve que 37:

37. The solution set of the equation ${\sin }^{-1}\sqrt{1-{\mathrm{x}}^2}+{\cos }^{-1}\mathrm{x}={\cos }^{-1} \left(\frac{\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}\right)-{\sin }^{-1}\mathrm{x}$
(A) [-1,1]-{0}                  (B) (0,1] $\cup${-1}          (C) [-1,0) $\cup${1}        (D) ​[-1,1]

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\sin }^{-1}\sqrt{1-{\mathrm{x}}^2}+{\cos }^{-1}\mathrm{x}={\cot }^{-1}\left(\frac{\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}\right)-{\sin }^{-1}\mathrm{x} \\ \mathrm{Set} {\cos }^{-1}\mathrm{x}=\mathrm{A} \\ \mathrm{As} \mathrm{range} \mathrm{of} {\cos }^{-1}\mathrm{x} \mathrm{is} \left[0,\mathrm{\pi }\right] \\ \mathrm{Hence} \mathrm{A}\in \left[0,\mathrm{\pi }\right] \\ \mathrm{x}=\mathrm{cosA}, \mathrm{equation} \mathrm{becomes} \\ {\sin }^{-1}\sqrt{1-{\cos }^2\mathrm{A}}+{\cos }^{-1}\left(\mathrm{cosA}\right)={\cot }^{-1}\left(\frac{\sqrt{1-{\cos }^2\mathrm{A}}}{\mathrm{cosA}}\right)-{\sin }^{-1}\left(\mathrm{cosA}\right) \\ \Rightarrow {\sin }^{-1}\sqrt{{\sin }^2\mathrm{A}}+\mathrm{A}={\cot }^{-1}\left(\frac{\sqrt{{\sin }^2\mathrm{A}}}{\mathrm{cosA}}\right)-\left[\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\left(\mathrm{cosA}\right)\right] \\ \left[\mathrm{As} {\sin }^{-1}\mathrm{\alpha }+{\cos }^{-1}\mathrm{\alpha }=\frac{\mathrm{\pi }}{2}\right] \mathrm{and} \left[{\cos }^{-1}\left(\mathrm{cos\alpha }\right)=\mathrm{\alpha } \mathrm{when} \mathrm{\alpha }\in \left[0,\mathrm{\pi }\right]\right] \\ \Rightarrow {\sin }^{-1}\left|\mathrm{sinA}\right|+\mathrm{A}={\cot }^{-1}\left(\frac{\left|\mathrm{sinA}\right|}{\mathrm{cosA}}\right)-\left[\frac{\mathrm{\pi }}{2}-\mathrm{A}\right] \\ \mathrm{As} \mathrm{sinA} \mathrm{will} \mathrm{be} \mathrm{non}-\mathrm{negative} \mathrm{when} \mathrm{A}\in \left[0,\mathrm{\pi }\right], \mathrm{hence} \left|\mathrm{sinA}\right|=\mathrm{sinA} \\ \Rightarrow {\sin }^{-1}\left(\mathrm{sinA}\right)+\mathrm{A}={\cot }^{-1}\left(\frac{\mathrm{sinA}}{\mathrm{cosA}}\right)-\frac{\mathrm{\pi }}{2}+\mathrm{A} \\ \Rightarrow {\sin }^{-1}\left(\mathrm{sinA}\right)={\cot }^{-1}\left(\mathrm{tanA}\right)-\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\sin }^{-1}\left(\mathrm{sinA}\right)=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\mathrm{tanA}\right)-\frac{\mathrm{\pi }}{2} \\ \left[\mathrm{As} {\cot }^{-1}\mathrm{\alpha }+{\tan }^{-1}\mathrm{\alpha }=\frac{\mathrm{\pi }}{2}\right] \\ \Rightarrow {\sin }^{-1}\left(\mathrm{sinA}\right)=-{\tan }^{-1}\left(\mathrm{tanA}\right) \\ \mathrm{Case} 1:\mathrm{When} \mathrm{A}\in [0,\frac{\mathrm{\pi }}{2}) \\ \mathrm{A}=-\mathrm{A}\backslash \\ \left[\mathrm{As} {\sin }^{-1}\left(\mathrm{sinA}\right)=\mathrm{A} \mathrm{and} {\tan }^{-1}\left(\mathrm{tanA}\right)=\mathrm{A} \mathrm{when} \mathrm{A}\in [0,\frac{\mathrm{\pi }}{2})\right] \\ \Rightarrow 2\mathrm{A}=0 \\ \Rightarrow \mathrm{A}=0 \\ \mathrm{x}=\mathrm{cosA}=\cos 0=1 \\ \mathrm{Case} 2:\mathrm{When} \mathrm{A}\in \left[\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right] \\ \Rightarrow \mathrm{\pi }-\mathrm{A}=-\left(\mathrm{A}-\mathrm{\pi }\right) \\ \left[ \mathrm{As} {\sin }^{-1}\left(\mathrm{sinA}\right)=\mathrm{\pi }-\mathrm{A} \mathrm{and} {\tan }^{-1}\left(\mathrm{tanA}\right)=\mathrm{A}-\mathrm{\pi } \mathrm{when} \mathrm{A}\in \left[\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right]\right] \\ \Rightarrow \mathrm{\pi }-\mathrm{A}=\mathrm{\pi }-\mathrm{A} \mathrm{which} \mathrm{is} \mathrm{true} \\ \mathrm{Hence} \mathrm{A}\in \left[\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right] \\ \Rightarrow \mathrm{x}=\mathrm{cosA}\in \left[\mathrm{cos\pi },\cos \frac{\mathrm{\pi }}{2}\right] \\ \Rightarrow \mathrm{x}\in \left[-1,0\right] \\ \mathrm{But} \mathrm{x}=0 \mathrm{will} \mathrm{be} \mathrm{taken} \mathrm{because} {\cot }^{-1}\left(\frac{\sqrt{1-{\mathrm{x}}^2}}{\mathrm{x}}\right) \mathrm{is} \mathrm{not} \mathrm{defined} \\ \mathrm{Hence} \mathrm{x}\in [-1,0) \\ \mathrm{Combining} \mathrm{two} \mathrm{cases}, \mathrm{we} \mathrm{get} \mathbf{x}\mathbf{\in }\mathbf{[}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\cup }\left\{\mathbf{1}\right\} \end{gathered}$$

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