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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 $\frac{21}{8}$.