Solve this

Given: $\cos \theta=\frac{-\sqrt{3}}{2}$

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

We know that,

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

Putting the values, we get

$\left(-\frac{\sqrt{3}}{2}\right)^{2}+\sin ^{2} \theta=1$ [given]

$\Rightarrow \frac{3}{4}+\sin ^{2} \theta=1$

$\Rightarrow \sin ^{2} \theta=1-\frac{3}{4}$

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

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

$\Rightarrow \sin \theta=\sqrt{\frac{1}{4}}$

$\Rightarrow \sin \theta=\pm \frac{1}{2}$

Since, $\theta$ in III $^{\text {rd }}$ quadrant and $\sin \theta$ is negative in III $^{\text {rd }}$ quadrant

$\therefore \sin \theta=-\frac{1}{2}$

Now,

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

Putting the values, we get

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

$=-\frac{1}{2} \times\left(-\frac{2}{\sqrt{3}}\right)$

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

Now,

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

Putting the values, we get

$\operatorname{cosec} \theta=\frac{1}{-\frac{1}{2}}$

$=-2$

Now,

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

Putting the values, we get

$\sec \theta=\frac{1}{-\frac{\sqrt{3}}{2}}$

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

Now,

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

Putting the values, we get

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

$=\sqrt{3}$

Hence, the values of other trigonometric Functions are: