Question
If $\tan\theta=\frac{1}{\sqrt{2}},$ find the value of $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\cot^2\theta}.$
✓
Answer
Given, $\tan\theta=\frac{1}{\sqrt{2}}$
We have to find the value of the expression $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\cot^2\theta}.$
We know that,
$1+\cot^2\theta=\text{cosec}^2\theta$
$\Rightarrow\ \text{cosec}^2\theta-\cot^2\theta=1$
Therefore, the given expression can be written as
$\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\cot^2\theta}=\frac{\text{cosec}^2\theta-\sec^2\theta}{1+\cot^2\theta+\cot^2\theta}$
$=\frac{\text{cosec}^2\theta-\sec^2\theta}{1+2\cot^2\theta}$
$\tan\theta=\frac{1}{\sqrt{2}}\Rightarrow\ \cot\theta=\sqrt{2}$
$\frac{\text{cosec}^2\theta-\sec^2\theta}{1+2\cot^2\theta}=\frac{1+\cot^2\theta-(1+\tan^2\theta)}{1+2\cot^2\theta}$
$(\text{since }\sec^2\theta=1+\tan^2\theta)$
$=\frac{\cot^2\theta-\tan^2\theta}{1+2\cot^2\theta}$
$=\frac{(\sqrt{2})^2-\Big(\frac{1}{\sqrt{2}}\Big)^2}{1+2\times(\sqrt{2})^2}$
$=\frac{3}{10}$
Hence, the value of the given expression is $\frac{3}{10}$.
We have to find the value of the expression $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\cot^2\theta}.$
We know that,
$1+\cot^2\theta=\text{cosec}^2\theta$
$\Rightarrow\ \text{cosec}^2\theta-\cot^2\theta=1$
Therefore, the given expression can be written as
$\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\cot^2\theta}=\frac{\text{cosec}^2\theta-\sec^2\theta}{1+\cot^2\theta+\cot^2\theta}$
$=\frac{\text{cosec}^2\theta-\sec^2\theta}{1+2\cot^2\theta}$
$\tan\theta=\frac{1}{\sqrt{2}}\Rightarrow\ \cot\theta=\sqrt{2}$
$\frac{\text{cosec}^2\theta-\sec^2\theta}{1+2\cot^2\theta}=\frac{1+\cot^2\theta-(1+\tan^2\theta)}{1+2\cot^2\theta}$
$(\text{since }\sec^2\theta=1+\tan^2\theta)$
$=\frac{\cot^2\theta-\tan^2\theta}{1+2\cot^2\theta}$
$=\frac{(\sqrt{2})^2-\Big(\frac{1}{\sqrt{2}}\Big)^2}{1+2\times(\sqrt{2})^2}$
$=\frac{3}{10}$
Hence, the value of the given expression is $\frac{3}{10}$.
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