If 3 cos A = 4 sin A, find the value of :

(i) cos A

(ii) 3 - cot² A + cosec² A

Given:

3 cos A = 4 sin A

$\dfrac{\text{cos A}}{\text{sin A}} = \dfrac{4}{3}$

$\text{cot A} = \dfrac{4}{3}$

$\text{cot A} = \dfrac{Base}{Perpendicular} = \dfrac{4}{3}$

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

In Δ ABC,

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

⇒ AC² = (4x)² + (3x)²

⇒ AC² = 16x² + 9x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

(i) cos $A = \dfrac{Base}{Hypotenuse}$

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

Hence, cos $A = \dfrac{4}{5}$.

(ii) 3 - cot² A + cosec² A

cot $A = \dfrac{Base}{Perpendicular}$

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

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

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

Now,

3 - cot² A + cosec² A

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

Hence, 3 - cot² A + cosec² A = 4.