Given 4 sin θ = 3 cos θ, find the values of :

(i) sin θ

(ii) cos θ

(iii) cot² θ - cosec² θ

Let ABC be a triangle with ∠B = 90° and ∠C = θ.

Given,

⇒ 4 sin θ = 3 cos θ

⇒ $\dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{3}{4}$.

⇒ tan θ = $\dfrac{3}{4}$ ………..(1)

From figure.

⇒ tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC}$ …………(2)

From (1) and (2) we get :

$\dfrac{AB}{BC} = \dfrac{3}{4}$

Let AB = 3x and BC = 4x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC² = $\sqrt{25x^2}$

⇒ AC = 5x.

(i) By formula,

⇒ sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{3x}{5x} = \dfrac{3}{5}$.

Hence, sin θ = $\dfrac{3}{5}$.

(ii) By formula,

⇒ cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$.

Hence, cos θ = $\dfrac{4}{5}$.

(iii) By formula,

⇒ cot θ = $\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{BC}{AB} = \dfrac{4x}{3x} = \dfrac{4}{3}$.

⇒ cot² θ = $\Big(\dfrac{4}{3}\Big)^2$ = $\dfrac{16}{9}$.

⇒ cosec θ = $\dfrac{\text{Hypotenuse}}{\text{Perpendicular}} = \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3}$.

⇒ cosec² θ = $\Big(\dfrac{5}{3}\Big)^2$ = $\dfrac{25}{9}$.

$\text{cot}^2 \text{ θ} - \text{cosec}^2 \text{ θ} = \dfrac{16}{9} - \dfrac{25}{9} \\[1em] = \dfrac{16 - 25}{9} \\[1em] = -\dfrac{9}{9} \\[1em] = -1.$

Hence, cot² θ - cosec² θ = -1.