Given : $\text{tan A} = \dfrac{4}{3}$, find : $\dfrac{\text{cosec A}}{\text{cot A} - \text{sec A}}$

Given:

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

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

∴ 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

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

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

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

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

sec $A = \dfrac{Hypotenuse}{Base}$

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

Now,

$\dfrac{\text{cosec A}}{\text{cot A} - \text{sec A}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{3}{4} - \dfrac{5}{3}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{3 \times 3}{4 \times 3} - \dfrac{5 \times 4}{3 \times 4}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{9}{12} - \dfrac{20}{12}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{9 - 20}{12}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{- 11}{12}}\\[1em] = \dfrac{5 \times 12}{- 11 \times 4}\\[1em] = \dfrac{- 60}{44}\\[1em] = \dfrac{- 15}{11}$

Hence, $\dfrac{\text{cosec A}}{\text{cot A} - \text{sec A}} = \dfrac{- 15}{11}$.