• question_answer Solve ${{\sin }^{-1}}x+{{\sin }^{-1}}(1-x)={{\cos }^{-1}}x.$

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 $x=\frac{1}{2}$