Question

Find the value(s) of x which satisfy the equation:
${\sin }^{-1}x+{\sin }^{-1}(1-x)={\cos }^{-1}x.$

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. No soltuion

Answer 1

Answer: (A), (B)

The correct options are
A $0$
B $\frac{1}{2}$

${\sin }^{-1}x+{\sin }^{-1}\left(1-x\right)={\cos }^{-1}x\mathrm{From}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{have}\sin \left({\sin }^{-1}x+{\sin }^{-1}\left(1-x\right)\right)=\sin \left({\cos }^{-1}x\right)\sin \left({\sin }^{-1}x\right)\cos \left({\sin }^{-1}\left(1-x\right)\right)+\cos \left({\sin }^{-1}x\right)\sin \left({\sin }^{-1}\left(1-x\right)\right)=\sin \left({\cos }^{-1}x\right)\Rightarrow x\sqrt{1-{\left(1-x\right)}^2}+\left(1-x\right)\sqrt{1-x^2}=\sqrt{1-x^2}\Rightarrow x\sqrt{2x-x^2}+\sqrt{1-x^2}\left(1-x-1\right)=0\Rightarrow x\left(\sqrt{2x-x^2}-\sqrt{1-x^2}\right)=0\Rightarrow x=0or2x-x^2=1-x^2\Rightarrow x=0orx=\frac{1}{2}.$