Solve this

Given: $\sin x=-\frac{2 \sqrt{6}}{5}$

To find: cos x and cot x

Since, $x$ is in IIIrd Quadrant. So, sin and cos will be negative but tan will be positive.

We know that,

$\sin ^{2} x+\cos ^{2} x=1$

Putting the values, we get

$\left(-\frac{2 \sqrt{6}}{5}\right)^{2}+\cos ^{2} x=1$ [Given]

$\Rightarrow \frac{24}{25}+\cos ^{2} x=1$

$\Rightarrow \cos ^{2} x=1-\frac{24}{25}$

$\Rightarrow \cos ^{2} x=\frac{25-24}{25}$

$\Rightarrow \cos ^{2} x=\frac{1}{25}$

$\Rightarrow \cos x=\sqrt{\frac{1}{25}}$

$\Rightarrow \cos x=\pm \frac{1}{5}$

Since, $x$ in IIIrd quadrant and $\cos x$ is negative in IIIrd quadrant

$\therefore \cos x=-\frac{1}{5}$

Now

$\tan x=\frac{\sin x}{\cos x}$

Putting the values, we get

$\tan x=\frac{-\frac{2 \sqrt{6}}{5}}{-\frac{1}{5}}$

$=-\frac{2 \sqrt{6}}{5} \times(-5)$

$=2 \sqrt{6}$

Now,

$\cot x=\frac{1}{\tan x}$

Putting the values, we get

$\cot x=\frac{1}{2 \sqrt{6}}$

Hence, the values of other trigonometric Functions are: