Solve this

Solution:

Given: $\sin \theta=\frac{-1}{2}$

Since, $\theta$ is in IVth Quadrant. So, sin and tan will be negative but cos will be positive.

We know that,

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

Putting the values, we get

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

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

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

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

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

$\Rightarrow \cos \theta=\sqrt{\frac{3}{4}}$

$\Rightarrow \cos \theta=\pm \frac{\sqrt{3}}{2}$

Since, $\theta$ in IV th quadrant and $\cos \theta$ is positive in IV $^{\text {th }}$ quadrant

$\therefore \cos \theta=\frac{\sqrt{3}}{2}$

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: