Solve this

Solution:

Given: $\sec \theta=\sqrt{2}$

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

Now, we know that

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

Putting the values, we get

$\cos \theta=\frac{1}{\sqrt{2}}$ …(i)

We know that,

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

Putting the values, we get

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

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

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

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

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

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

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

Since, $\theta$ in IV $^{\text {th }}$ quadrant and $\sin \theta$ is negative in IV $^{\text {th }}$ quadrant

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

Now,

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

Putting the values, we get

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

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

$=-1$

Now,

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

Putting the values, we get

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

$=-\sqrt{2}$

Now

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

Putting the values, we get

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

$=-1$

Hence, the values of other trigonometric Functions are: