Find value of k if sin^-1 ≤ft[ k tan ≤ft( 2 cos^-1 (√3)/2 ) ] = (π)/3.

The correct answer is Option (2) → $\frac{1}{2}$ ##

We have,

$\sin^{-1} \left[ k \tan \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right] = \frac{\pi}{3}$

$k \left[ \tan \left( 2 \cos^{-1} \left( \cos \frac{\pi}{6} \right) \right) \right] = \sin \frac{\pi}{3}$

$k \tan \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$

$k \sqrt{3} = \frac{\sqrt{3}}{2} ⇒k = \frac{1}{2}$