Question

If cosec θ = $\sqrt{10}$, then sec θ = ?
(a) $\frac{3}{\sqrt{10}}$
(b) $\frac{\sqrt{10}}{3}$
(c) $\frac{1}{\sqrt{10}}$
(d) $\frac{2}{\sqrt{10}}$

Answer 1

(b) $\frac{\sqrt{10}}{3}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: cosec $\theta$ = $\sqrt{10}$, but sin $\theta$ = $\frac{1}{\cos ec\theta }$ = $\frac{1}{\sqrt{10}}$
Also, sin $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}$ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{1}{\sqrt{10}}$
Thus, BC = k and AC = $\sqrt{10}$ k


Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = ($\sqrt{10}$ k)² $-$ (k)²
⇒ AB² = 9k²
⇒ AB = 3k
∴ sec $\theta$ = $\frac{AC}{AB}$ = $\frac{\sqrt{10}k}{3k}=\frac{\sqrt{10}}{3}$