Question

If tan θ = $\frac{8}{15}$, then cosec θ = ?
(a) $\frac{17}{8}$
(b) $\frac{8}{17}$
(c) $\frac{17}{15}$
(d) $\frac{15}{17}$

Answer 1

(a) $\frac{17}{8}$

Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: tan $\theta$ = $\frac{8}{15}$, but tan $\theta$ = $\frac{BC}{AB}$
So, $\frac{BC}{AB}$ = $\frac{8}{15}$
Thus, BC = 8k and AB = 15k


Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AC² = (15k)² + (8k)²
⇒ AC² = 289k²
⇒ AC = 17k
∴ cosec $\theta$ = $\frac{AC}{BC}=\frac{17k}{8k}=\frac{17}{8}$