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