Question : Find the value of the given expression.
$\frac{4}{3} \tan^2 45^{\circ}+3 \cos^2 30^{\circ}-2 \sec^2 30^{\circ}-\frac{3}{4} \cot^2 60^{\circ}$
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{\sqrt{2}}{3}$
Option 4: $\frac{3}{\sqrt{2}}$
Correct Answer: $\frac{2}{3}$
Solution :
Given: $\frac{4}{3} \tan^2 45^{\circ}+3 \cos^2 30^{\circ}-2 \sec^2 30^{\circ}-\frac{3}{4} \cot^2 60^{\circ}$
$=\frac{4}{3} \times 1 +3 ( \frac{\sqrt3}{2})^2-2 (\frac{2}{\sqrt3})^2-\frac{3}{4} (\frac{1}{\sqrt3})^2$
$= \frac{4}{3} \times 1^2 +\frac{9}4-\frac{8}3-\frac{3}{4}\times \frac{1}3$
$= \frac{4}3+\frac{9}4-\frac{8}3-\frac{1}4$
$= \frac{16+27-32-3}{12}$
$= \frac{2}3$
Hence, the correct answer is $\frac{2}3$.
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