Answer
$0$
Work Step by Step
Let us consider $p=\sin^{-1} \dfrac{\sqrt 3}{2}$ and $q=\cot^{-1} {\sqrt 3}$
This gives $p=\dfrac{\pi}{3}$ and $q=\dfrac{\pi}{6}$
Now, $\cos(\sin^{-1} \dfrac{\sqrt 3}{2}+\cot^{-1} {\sqrt 3})=\cos (p+q)$
This implies that
$\cos (p+q)=\cos (\dfrac{\pi}{3}+\dfrac{\pi}{6})=\cos \dfrac{\pi}{2}$
But $\cos \dfrac{\pi}{2}=0$
Hence, $\cos(\sin^{-1} \dfrac{\sqrt 3}{2}+\cot^{-1} {\sqrt 3})=0$
