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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 $\frac{3}{10}$.