If cot θ=13√, write the value of 1−cos2 θ2−sin2 θ.

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 $\frac{3}{5}$