If $cos^{-1}\sqrt{3}x+cos^{-1}x=\frac{\pi}{2}$, then the value of x is :

$\frac{1}{2}$ $-\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $-\frac{1}{\sqrt{2}}$

$\frac{1}{2}$

The correct answer is Option (1) → $\frac{1}{2}$ $\cos^{-1}\sqrt{3}x+\cos^{-1}x=\frac{\pi}{2}$ $\cos^{-1}\sqrt{3}x+\cos^{-1}x=\sin^{-1}x+\cos^{-1}x$ $\cos^{-1}\sqrt{3}x=\sin^{-1}x$ so $(\sqrt{3}x)^2=1-x^2$ $x^2=\frac{1}{4}$ $x=±\frac{1}{2}⇒x=\frac{1}{2}$ as $-\frac{1}{2}$ doesn't satisfied equation