In a right-angled triangle, it is given that angle A is an acute angle and that tan A = $\dfrac{5}{12}$. Find the values of :

(i) cos A

(ii) cosec A - cot A.

Let ABC be a right angled triangle with ∠B = 90°.

Given,

tan A = $\dfrac{5}{12}$.

By formula,

tan A = $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{BC}{AB} = \dfrac{5}{12}$

Let BC = 5x and AB = 12x.

From right-angled ∆ABC

By pythagoras theorem, we get

⇒ AC² = BC² + AB²

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169x^2}$

⇒ AC = 13x.

(i) By formula,

cos A = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{AB}{AC} = \dfrac{12x}{13x} = \dfrac{12}{13}$.

Hence, cos A = $\dfrac{12}{13}$.

(ii) By formula,

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

= $\dfrac{AC}{BC} = \dfrac{13x}{5x} = \dfrac{13}{5}$.

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

= $\dfrac{AB}{BC}$ = $\dfrac{12x}{5x}$ = $\dfrac{12}{5}$.

Substituting values in cosec A - cot A, we get :

$\Rightarrow \text{cosec A - cot A} = \dfrac{13}{5} - \dfrac{12}{5} \\[1em] = \dfrac{1}{5}.$

Hence, cosec A - cot A = $\dfrac{1}{5}$.