# If $\cot \theta=\frac{1}{\sqrt{3}}$, find the value of $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$

Given:

$\cot \theta=\frac{1}{\sqrt{3}}$

To do:

We have to evaluate $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$.

Solution:

We know that,

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

$\sin ^{2} \theta=\frac{1}{\operatorname{cosec} ^{2} \theta}$

$=\frac{1}{1+\cot ^{2} \theta}$
$\Rightarrow \sin \theta=\frac{1}{\sqrt{1+\cot ^{2} \theta}}$

$=\frac{1}{\sqrt{1+(\frac{1}{\sqrt3})^2}}$

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

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

$=\frac{1}{\sqrt{\frac{4}{3}}}$

$=\frac{1}{\frac{2}{\sqrt3}}$

$=\frac{\sqrt3}{2}$

$\cos \theta=\sqrt{1-\sin^2 \theta}$

$=\sqrt{1-(\frac{\sqrt3}{2})^2}$

$=\sqrt{\frac{4-3}{4}}$

$=\sqrt{\frac{1}{4}}$

$=\frac{1}{2}$

This implies,

$\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{1-(\frac{1}{2})^2}{2-(\frac{\sqrt3}{2})^2}$

$=\frac{\frac{4-1}{4}}{\frac{2(4)-3}{4}}$

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

$=\frac{3}{5}$

The value of $\frac{1- \cos^2 \theta}{2- \sin^2 \theta}$ is $\frac{3}{5}$.

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Updated on: 10-Oct-2022

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