Solve this

Solution:

Given: $\sec x=-2$

Given that: $\pi

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

Now, we know that

$\cos x=\frac{1}{\sec x}$

Putting the values, we get

$\cos x=\frac{1}{-2}$ …(i)

We know that,

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

Putting the values, we get

$\left(-\frac{1}{2}\right)^{2}+\sin ^{2} x=1$ [Given]

$\Rightarrow \frac{1}{4}+\sin ^{2} x=1$

$\Rightarrow \sin ^{2} x=1-\frac{1}{4}$

$\Rightarrow \sin ^{2} x=\frac{4-1}{4}$

$\Rightarrow \sin ^{2} x=\frac{3}{4}$

$\Rightarrow \sin x=\sqrt{\frac{3}{4}}$

$\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$

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

$\therefore \sin x=-\frac{\sqrt{3}}{2}$

Now,

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

Putting the values, we get

$\tan x=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$

$=-\frac{\sqrt{3}}{2} \times(-2)$

$=\sqrt{3}$

Now,

$\operatorname{cosec} x=\frac{1}{\sin x}$

Putting the values, we get

$\operatorname{cosec} x=\frac{1}{-\frac{\sqrt{3}}{2}}$

$=-\frac{2}{\sqrt{3}}$

Now

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

Putting the values, we get

$\cot x=\frac{1}{\sqrt{3}}$

Hence, the values of other trigonometric Functions are: