If θ is an acute angle and tan θ = $\dfrac{8}{15}$, find the value of sec θ + cosec θ.

Given,

tan θ = $\dfrac{8}{15}$ ……….(1)

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

By formula,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC}$ ………(2)

From (1) and (2) we get,

$\dfrac{AB}{BC} = \dfrac{8}{15}$

Let AB = 8x and BC = 15x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (8x)² + (15x)²

⇒ AC² = 64x² + 225x²

⇒ AC² = 289x²

⇒ AC = $\sqrt{289x^2}$

⇒ AC = 17x.

By formula,

sec θ = $\dfrac{\text{Hypotenuse}}{\text{Base}}$

= $\dfrac{AC}{BC} = \dfrac{17x}{15x} = \dfrac{17}{15}$.

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

= $\dfrac{AC}{AB} = \dfrac{17x}{8x} = \dfrac{17}{8}$.

Substituting value in sec θ + cosec θ we get :

$\Rightarrow \text{sec θ + cosec θ} = \dfrac{17}{15} + \dfrac{17}{8} \\[1em] = \dfrac{136 + 255}{120} \\[1em] = \dfrac{391}{120} \\[1em] = 3\dfrac{31}{120}.$

Hence, sec θ + cosec θ = $3\dfrac{31}{120}$.