If $\text{cosec θ} = {\sqrt5}$ find the value of :

(i) 2 - sin² θ - cos² θ

(ii) $2 + \dfrac{1}{\text{sin}^2 \text{ θ}} - \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$

Given:

$\text{cosec θ} = {\sqrt5}$

$⇒ \text{cosec θ} = \dfrac{Hypotenuse}{Perpendicular} = {\sqrt5}\\[1em]$

∴ If length of AC = $\sqrt{5}$x unit, length of BC = x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ ($\sqrt{5}$x)² = (x)² + AB²

⇒ 5x² = x² + AB²

⇒ AB² = 5x² - x²

⇒ AB² = 4x²

⇒ AB = $\sqrt{4 \text{x}^2}$

⇒ AB = 2x

(i) sin θ = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{CA} = \dfrac{x}{\sqrt{5}x} = \dfrac{1}{\sqrt{5}}$

cos θ = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{2x}{\sqrt{5}x} = \dfrac{2}{\sqrt{5}}$

Now,

2 - sin²θ - cos²θ

$= 2 - \Big(\dfrac{1}{\sqrt{5}}\Big)^2 - \Big(\dfrac{2}{\sqrt{5}}\Big)^2\\[1em] = 2 - \dfrac{1}{5} - \dfrac{4}{5}\\[1em] = 2 + \dfrac{-1 - 4}{5}\\[1em] = 2 + \dfrac{-5}{5}\\[1em] = 2 - 1\\[1em] = 1$

Hence, 2 - sin²θ - cos²θ = 1.

(ii) $2 + \dfrac{1}{\text{sin}^2 \text{ θ}} - \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$

$= 2 + \dfrac{1}{\Big(\dfrac{1}{\sqrt{5}}\Big)^2} - \dfrac{\Big(\dfrac{2}{\sqrt{5}}\Big)^2}{\Big(\dfrac{1}{\sqrt{5}}\Big)^2}\\[1em] = 2 + \dfrac{1}{\dfrac{1}{5}} - \dfrac{\dfrac{4}{5}}{\dfrac{1}{5}}\\[1em] = 2 + \dfrac{5}{1} - \dfrac{\dfrac{4}{\cancel{5}}}{\dfrac{1}{\cancel{5}}}\\[1em] = 2 + 5 - \dfrac{4}{1}\\[1em] = 7 - 4\\[1em] = 3$

Hence, $2 + \dfrac{1}{\text{sin}^2 \text{ θ}} - \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$ = 3.