If tan θ=12√, find the value of cosec2 θ−sec2 θcosec2 θ+cot2 θ.

Question:

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

Solution:

Given: $\tan \theta=\frac{1}{\sqrt{2}}$

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

We know that,

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

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

Therefore, the given expression can be written as

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

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

$\tan \theta=\frac{1}{\sqrt{2}} \Rightarrow \cot \theta=\sqrt{2}$

$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{1+2 \cot ^{2} \theta}=\frac{1+\cot ^{2} \theta-\left(1+\tan ^{2} \theta\right)}{1+2 \cot ^{2} \theta} \quad\left(\right.$ since $\left.\sec ^{2} \theta=1+\tan ^{2} \theta\right)$

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

$=\frac{(\sqrt{2})^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}}{1+2 \times(\sqrt{2})^{2}}$

$=\frac{3}{10}$

Hence, the value of the given expression is $\frac{3}{10}$.