If 5 cos θ = 3, evaluate :

$\dfrac{\text{cosec θ} - \text{cot θ}}{\text{cosec θ} + \text{cot θ}}$.

Given:

5 cos θ = 3

$\text{cos θ} = \dfrac{3}{5}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{3}{5}$

∴ If length of BA = 3x unit, length of CA = 5x unit.

In Δ ABC,

⇒ AC² = AB² + CB² (∵ AC is hypotenuse)

⇒ (5x)² = (3x)² + CB²

⇒ 25x² = 9x² + CB²

⇒ CB² = 25x² - 9x²

⇒ CB² = 16x²

⇒ CB = $\sqrt{16\text{x}^2}$

⇒ CB = 4x

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

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

cot θ = $\dfrac{Base}{Perpendicular}$

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

Now,

$\dfrac{\text{cosec θ} - \text{cot θ}}{\text{cosec θ} + \text{cot θ}}\\[1em] = \dfrac{\dfrac{5}{4} - \dfrac{3}{4}}{\dfrac{5}{4} + \dfrac{3}{4}}\\[1em] = \dfrac{\dfrac{5 - 3}{4}}{\dfrac{5 + 3}{4}}\\[1em] = \dfrac{\dfrac{2}{4}}{\dfrac{8}{4}}\\[1em] = \dfrac{\dfrac{2}{\cancel{4}}}{\dfrac{8}{\cancel{4}}}\\[1em] = \dfrac{2}{8}\\[1em] = \dfrac{1}{4}$

Hence, $\dfrac{\text{cosec θ} - \text{cot θ}}{\text{cosec θ} + \text{cot θ}} = \dfrac{1}{4}$.