Given : 4 cot A = 3, find :

(i) sin A

(ii) sec A

(iii) cosec² A - cot² A

Given:

4 cot A = 3

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

i.e., $\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{3}{4}$

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

In Δ ABC,

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

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

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

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

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

(ii) sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3} = 1\dfrac{2}{3}$

Hence, sec $A = \dfrac{5}{3} = 1\dfrac{2}{3}$.

(iii) cosec² A - cot² A

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

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

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

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

Now,

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

Hence, cosec² A - cot² A = 1.