Evaluate: $\frac{2 \cos 30^\circ - \cot^3 60^\circ}{\tan 30^\circ}$. (Q 22 B - 30/5/2 - CBSE 2026 Question Paper)

We know that cos 30 0 = $\frac{\sqrt 3}{2}$ cot 60 0 = $\frac{1}{\sqrt 3}$ and tan 30 0 = $\frac{1}{\sqrt 3}$ Given expression is: $\frac{2 \cos 30^\circ - \cot^3 60^\circ}{\tan 30^\circ}$ By substituting these values, we get: = $\frac{(2 (\cos 30^\circ) - (\cot 60^\circ)^3)}{(\tan 30^\circ)}$ = $\frac{2 (\frac{\sqrt 3}{2}) - (\frac{1}{\sqrt 3})^3}{\frac{1}{\sqrt 3}}$ = $\frac{(\sqrt 3) - (\frac{1}{3\sqrt 3})}{\frac{1}{\sqrt 3}}$ = $\frac{(9 - 1)}{3\sqrt 3} \times {\frac{\sqrt 3}{1}}$ = $\frac{8\sqrt 3}{3\sqrt 3}$ = $\frac{8}{3}$ Therefore, the value of the given expression is $\frac{8}{3}$. Please press "Heart" if you liked the solution.