If $\cot \theta=\frac{1}{\sqrt{3}}$, write the value of $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$.
Given: $\cot \theta=\frac{1}{\sqrt{3}}$
$\frac{\text { Base }}{\text { Perpendicular }}=\frac{1}{\sqrt{3}}$
Base $=1$
Perpendicular $=\sqrt{3}$
Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$
Hypotenuse $=2$
Now we find, $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$
$=\frac{1-\frac{(\text { Base })^{2}}{(\text { hypotenuse })^{2}}}{2-\frac{(\text { Perpendicular })^{2}}{(\text { hypotenuse })^{2}}}$
$=\frac{1-\frac{(1)^{2}}{(2)^{2}}}{2-\frac{(\sqrt{3})^{2}}{(2)^{2}}}$
$=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}$
$=\frac{3}{5}$
Hence the value of $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$ is $\frac{3}{5}$
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