# If $\cot \theta=\frac{1}{\sqrt{3}}$, show that $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{3}{5}$

Given:

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

To do:

We have to show that $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{3}{5}$.

Solution:

Let, in a triangle $ABC$ right-angled at $B$, $\ cot\ \theta = cot\ A = \frac{1}{\sqrt{3}}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$cot\ \theta=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(1)^2+(\sqrt3)^2$

$\Rightarrow AC^2=1+3$

$\Rightarrow AC=\sqrt{4}=2$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{\sqrt3}{2}$

$cos\ \theta=\frac{AB}{AC}=\frac{1}{2}$

Let us consider LHS,

$\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{1-\left(\frac{1}{2}\right)^{2}}{2-\left(\frac{\sqrt{3}}{2}\right)^{2}}$

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

$=\frac{\frac{4-1}{4}}{\frac{8-3}{4}}$

$=\frac{3}{5}$

$=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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