If $ \cot \theta=\sqrt{3} $, find the value of $ \frac{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta} $

Given:

\( \cot \theta=\sqrt{3} \)

To do:

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

Solution:  

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

$\Rightarrow \cot^{2} \theta=(\sqrt{3})^2=3$

$\tan ^{2} \theta=\frac{1}{\cot ^{2} \theta}$

$=\frac{1}{3}$
$\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta$

$=1+3$

$=4$
$\sec ^{2} \theta=1+\tan ^{2} \theta$

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

$=\frac{4}{3}$

Therefore,

$\frac{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}=\frac{4+3}{4-\frac{4}{3}}$

$=\frac{7}{\frac{12-4}{3}}$

$=\frac{7}{\frac{8}{3}}$

$=\frac{7 \times 3}{8}$

$=\frac{21}{8}$

The value of \( \frac{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta} \) is $\frac{21}{8}$.

Tutorialspoint

e

Updated on: 10-Oct-2022

45 Views

Kickstart Your Career

Get certified by completing the course

Get Started