Find the value of:

$$\left[\frac{4}{3} \tan^2 60^\circ + 3 \cos^2 30^\circ - 2 \sec^2 30^\circ - \frac{3}{4} \cot^2 60^\circ \right] \div \left[ \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ \right]$$​

Correct option is A

​Given:

$$\left[\frac{4}{3} \tan^2 60^\circ + 3 \cos^2 30^\circ - 2 \sec^2 30^\circ - \frac{3}{4} \cot^2 60^\circ \right] \div \left[ \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ \right]\\$$

Concept Used:

Solution:

​$$\begin{aligned}&\Rightarrow[ \frac{4}{3} \left( \sqrt{3} \right)^2 + 3 \left( \frac{\sqrt{3}}{2} \right)^2 - 2 \left( \frac{2}{\sqrt{3}} \right)^2 - \frac{3}{4} \left( \frac{1}{\sqrt{3}} \right)^2] \div [\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}- \frac{1}{2}\times\frac{1}{2}] \\[10pt]&\Rightarrow [\frac{4}{3} \times 3 + 3 \times \frac{3}{4} - 2 \times \frac{4}{3} - \frac{3}{4} \times \frac{1}{3}] \div [\frac{3}{4}- \frac{1}{4}] \\[10pt]&\Rightarrow [4 + \frac{9}{4} - \frac{8}{3} - \frac{1}{4}] \div [\frac{1}{2}] \\&\Rightarrow[ \frac{(48 + 27 - 32 - 3)}{12}] \div [\frac{1}{2}] \\ &\Rightarrow \frac{40}{12} \div \frac{1}{2}\\ &\Rightarrow \frac{40}{12}\times 2 = \frac{20}{3} = 6 \frac{2}{3}\end{aligned}$$​​