Question

$\frac{4}{3}{\cot }^2{30}^{\circ }+3{\sin }^2{60}^{\circ }-2{\text{cosec}}^2{60}^{\circ }-\frac{3}{4}{\tan }^2{30}^{\circ }$ is

• A. $1$
• B. $-\frac{20}{3}$
• C. $\frac{10}{3}$
• D. $5$

Answer 1

The correct option is C $\frac{10}{3}$

$\frac{4}{3}{\cot }^2{30}^{\circ }+3{\sin }^2{60}^{\circ }-2{\text{cosec}}^2{60}^{\circ }-\frac{3}{4}{\tan }^2{30}^{\circ }$
$=\frac{4}{3}(\sqrt{3}{)}^2+3(\frac{\sqrt{3}}{2}{)}^2-2(\frac{2}{\sqrt{3}}{)}^2-\frac{3}{4}(\frac{1}{\sqrt{3}}{)}^2$
$=\frac{4}{3}\times 3+3\times \frac{3}{4}-2\times \frac{4}{3}-\frac{3}{4}\times \frac{1}{3}$
$=4+\frac{9}{4}-\frac{8}{3}-\frac{1}{4}$
$=\frac{10}{3}$
Option C is correct.