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

Find the values of :

(i) cos A

(ii) sin A

(iii) $\dfrac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}}$

Given:

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

i.e., $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{5}{12}$

∴ If length of BC = 5x unit, length of AB = 12x unit.

In Δ ABC,

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

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169\text{x}^2}$

⇒ AC = 13x

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

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

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

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

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

Hence, sin $A = \dfrac{5}{13}$.

(iii) $\dfrac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}}$

$= \dfrac{\dfrac{12}{13} + \dfrac{5}{13}}{\dfrac{12}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{12 + 5}{13}}{\dfrac{12 - 5}{13}}\\[1em] = \dfrac{\dfrac{17}{\cancel{13}}}{\dfrac{7}{\cancel{13}}}\\[1em] = \dfrac{17}{7}\\[1em] = 2\dfrac{3}{7}$

Hence, $\dfrac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}} = 2\dfrac{3}{7}$.