Given : 4 sin θ = 3 cos θ; find the value of :

(i) sin θ

(ii) cos θ

(iii) cot² θ - cosec² θ

(iv) 4 cos² θ - 3 sin² θ + 2

Given:

4 sin θ = 3 cos θ

$⇒ \dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{3}{4}\\[1em] ⇒ \text{tan θ} = \dfrac{3}{4}\\[1em] ⇒ \text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}}\\[1em] ⇒ \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{3}{4}\\[1em]$

∴ If length of BC = 3x unit, length of AB = 4x unit.

In Δ ABC,

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

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25 \text{x}^2}$

⇒ AC = 5x

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

$\dfrac{CB}{AC} = \dfrac{3x}{5x} = \dfrac{3}{5}$

Hence, sin θ = $\dfrac{3}{5}$.

(ii) cos θ = $\dfrac{Base}{Hypotenuse}$

$\dfrac{AB}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

Hence, cos θ = $\dfrac{4}{5}$.

(iii) cot² θ - cosec² θ + 2

cot θ = $\dfrac{Base}{Perpendicular}$

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

cosec θ = $\dfrac{Hypotenuse}{Perpendicular}$

$\dfrac{AC}{CB} = \dfrac{5x}{3x} = \dfrac{5}{3}$

Now, cot² θ - cosec² θ

$= \Big(\dfrac{4}{3}\Big)^2 - \Big(\dfrac{5}{3}\Big)^2 \\[1em] = \dfrac{16}{9} - \dfrac{25}{9} \\[1em] = \dfrac{16 - 25}{9} \\[1em] = \dfrac{- 9}{9} \\[1em] = -1$

Hence, cot² θ - cosec² θ + 2 = -1.

(iv) 4 cos² θ - 3 sin² θ + 2

cos θ = $\dfrac{Base}{Hypotenuse}$

$\dfrac{AB}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

sin θ = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{CB}{AC} = \dfrac{3x}{5x} = \dfrac{3}{5}$

Now, 4 cos² θ - 3 sin² θ + 2

$= 4 \times \Big(\dfrac{4}{5}\Big)^2 - 3 \times \Big(\dfrac{3}{5}\Big)^2 + 2\\[1em] = 4 \times \dfrac{16}{25} - 3 \times \dfrac{9}{25} + 2\\[1em] = \dfrac{64}{25} - \dfrac{27}{25} + 2\\[1em] = \dfrac{64 - 27}{25} + 2\\[1em] = \dfrac{37}{25} + 2\\[1em] = \dfrac{37}{25} + \dfrac{2 \times 25}{25}\\[1em] = \dfrac{37}{25} + \dfrac{50}{25}\\[1em] = \dfrac{37 + 50}{25}\\[1em] = \dfrac{87}{25}\\[1em] = 3\dfrac{12}{25}$

Hence, 4 cos² θ - 3 sin² θ + 2 =$3\dfrac{12}{25}$.