If $ \tan \theta=\frac{1}{\sqrt{2}} $, find the value of $ \frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta} $

Given:

\( \tan \theta=\frac{1}{\sqrt{2}} \)

To do:

We have to find the value of \( \frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta} \).

Solution:  

We know that,

$\sec ^{2} \theta-\tan ^{2} \theta=1$

$\operatorname{cosec} ^{2} \theta-\cot ^{2} \theta=1$

Therefore,

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

$=\frac{1}{\frac{1}{\sqrt{2}}}$

$=\sqrt{2}$ 

$\sec ^{2} \theta-\tan ^{2} \theta=1$

$\sec ^{2} \theta-(\frac{1}{\sqrt{2}})^2=1$

$\sec \theta=\sqrt{1+\frac{1}{2}}$

$=\sqrt{\frac{2+1}{2}}$

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

$\operatorname{cosec} ^{2} \theta-(\sqrt{2})^2=1$

$\operatorname{cosec} \theta=\sqrt{1+2}$

$=\sqrt{3}$

This implies,

$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta}=\frac{(\sqrt{3})^{2}-\left(\sqrt{\frac{3}{2}}\right)^{2}}{(\sqrt{3})^{2}+(\sqrt{2})^{2}}$

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

$=\frac{\frac{3}{2}}{5}$

$=\frac{3}{2\times 5}$

$=\frac{3}{10}$

The value of \( \frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta} \) is $\frac{3}{10}$.

Tutorialspoint

e

Updated on: 10-Oct-2022

45 Views

Kickstart Your Career

Get certified by completing the course

Get Started