Given A is an acute angle and cosec A = $\sqrt{2}$, find the value of

$\dfrac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A}$

Let triangle ABC be a right-angled at B and A is a acute angle.

Given,

cosec A = $\sqrt{2}$

⇒ cosec A = $\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\dfrac{AC}{BC} = \dfrac{\sqrt{2}}{1}$

Let AC = $\sqrt{2}$k and BC = k.

In right angled triangle ABC,

By using pythagoras theorem,

AC² = AB² + BC²

($\sqrt{2}k$)² = AB² + k²

2k² = AB² + k²

AB² = 2k² - k²

AB² = k²

AB = $\sqrt{\text{k}}$ = k.

By formula,

sin A = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{BC}{AC} = \dfrac{k}{\sqrt{2}k} = \dfrac{1}{\sqrt{2}}$.

tan A = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{BC}{AB} = \dfrac{k}{k} = 1$.

cot A = $\dfrac{1}{\text{tan A}} = \dfrac{1}{1} = 1$.

cos A = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{AB}{AC} = \dfrac{k}{\sqrt{2}k} = \dfrac{1}{\sqrt{2}}$.

Substituting these values in $\dfrac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A}$ we get,

$\Rightarrow \dfrac{2 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 + 3 \times 1^2}{1^2 - \Big(\dfrac{1}{\sqrt{2}}\Big)^2} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{2} + 3 \times 1}{1 - \dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{1 + 3}{\dfrac{1}{2}} \\[1em] \Rightarrow 4 \times 2 \\[1em] \Rightarrow 8.$

Hence, $\dfrac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A}$ = 8.