Answer:
We have, \[{{\sin }^{-1}}x+{{\sin }^{-1}}(1-x)={{\cos }^{-1}}x\] \[\Rightarrow \] \[{{\sin }^{-1}}x-{{\cos }^{-1}}x=-{{\sin }^{-1}}(1-x)\] \[\Rightarrow \] \[{{\sin }^{-1}}x-{{\cos }^{-1}}x={{\sin }^{-1}}(x-1)\] ?(i) \[[\because {{\sin }^{-1}}(-x)=-{{\sin }^{-1}}x]\] Put \[{{\sin }^{-1}}x=\alpha \]and \[{{\cos }^{-1}}x=\beta \] \[\Rightarrow \] \[\sin \alpha =x\]and \[\cos \beta =x\] Then, \[\cos \alpha =\sqrt{1-{{\sin }^{2}}\alpha }\]and \[\sin \beta =\sqrt{1-{{\cos }^{2}}\beta }\] \[\cos \alpha =\sqrt{1-{{x}^{2}}}\]and \[\sin \beta =\sqrt{1-{{x}^{2}}}\] Now, \[\sin (\alpha -\beta )=sin\alpha \cdot cos\beta -cos\alpha \cdot \sin \beta \] \[=x\cdot x-\sqrt{1-{{x}^{2}}}\sqrt{1-{{x}^{2}}}\] \[={{x}^{2}}-(1-{{x}^{2}})={{x}^{2}}-1+{{x}^{2}}\] \[\Rightarrow \]\[\sin (\alpha -\beta )=2{{x}^{2}}-1\Rightarrow \alpha -\beta ={{\sin }^{-1}}(2{{x}^{2}}-1)\] \[\Rightarrow \] \[{{\sin }^{-1}}x-{{\cos }^{-1}}x={{\sin }^{-1}}(2{{x}^{2}}-1)\] ?(ii) Now, from Eqs. (i) and (ii), we get \[{{\sin }^{-1}}(2{{x}^{2}}-1)={{\sin }^{-1}}(x-1)\Rightarrow 2{{x}^{2}}-1=x-1\] \[\Rightarrow \] \[2{{x}^{2}}-x=0\Rightarrow x\,(2x-1)=0\] \[\therefore \] \[x=0\]or \[x=\frac{1}{2}\]
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