If cot θ = $\dfrac{1}{\sqrt{3}}$, show that $\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}$.

cot θ = $\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{1}{\sqrt{3}}$

Let base = x and perpendicular = $\sqrt{3}$x

By using pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (x)² + $(\sqrt{3}x)^2$

Hypotenuse² = x² + 3x²

Hypotenuse² = 4x²

Hypotenuse = $\sqrt{4x^2}$

Hypotenuse = 2x

Now

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{\sqrt{3}x}{2x} = \dfrac{\sqrt{3}}{2}$

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{2x} = \dfrac{1}{2}$

Substituting values we get :

$\Rightarrow \Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{1 - \Big(\dfrac{1}{2}\Big)^2}{2 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2} \\[1em] = \dfrac{1 - \dfrac{1}{4}}{2 - \dfrac{3}{4}} \\[1em] = \dfrac{\dfrac{4 - 1}{4}}{\dfrac{8 - 3}{4}} \\[1em] = \dfrac{\dfrac{3}{4}}{\dfrac{5}{4}} \\[1em] = \dfrac{3}{5}.$

Hence, $\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}$.