Question

If $\sin x=\frac{-2\sqrt{6}}{5}$ and $x$ lies in III quadrant, then the value of $\cot x$ is

• A. $\frac{3}{\sqrt{6}}$
• B. $-\frac{3}{\sqrt{6}}$
• C. $\frac{1}{2\sqrt{6}}$
• D. $-\frac{1}{2\sqrt{6}}$

Answer 1

Answer: (C)

$\because {\cos }^{2}x=\left(1-{\sin }^{2}x\right)=\left(1-\frac{24}{25}\right)=\frac{1}{25}$
( $\because \sin x$ and $\cos x$ are negative in III quadrant)
$\Rightarrow \cos x=-\frac{1}{5}$
$\therefore \cot x=\frac{\cos x}{\sin x}=\frac{-1/5}{-2\sqrt{6}/5}=\frac{1}{2\sqrt{6}}$