The principal value of $cosec^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is :

$\frac{\pi}{3}$

The correct answer is Option (1) → $\frac{\pi}{3}$

$θ=cosec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Now,

$cosec\,θ=\frac{1}{\sin θ}$

and,

$\frac{1}{\sin θ}=\frac{2}{\sqrt{3}}⇒\sin θ=\frac{\sqrt{3}}{2}$

Now,

$\sin θ=\frac{\sqrt{3}}{2}$

$⇒θ=\frac{\pi}{3}$