Question

If $\sin A=\frac{\sqrt{3}}{2}$ and A is an acute angle, then find the value of $\frac{\tan A-\cot A}{\sqrt{3}+co\sec A}$

• A. $\frac{-2}{5}$
• B. $\frac{2}{5}$
• C. $\frac{2}{3+2\sqrt{3}}$
• D. $-2$

Answer 1

The correct option is B $\frac{2}{5}$

Given, $\sin A=\frac{\sqrt{3}}{2}$
$\Rightarrow A={60}^0$ (Since, $\sin {60}^0=\frac{\sqrt{3}}{2}$ and $0^0<A<{90}^0)$
So, $\frac{\tan A-\cot A}{\sqrt{3}+\text{cosec}A}=\frac{\tan {60}^0-\cot {60}^0}{\sqrt{3}+\text{cosec}{60}^0}=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{2}{\sqrt{3}}}=\frac{3-1}{3+2}=\frac{2}{5}$