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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.