Question

If sin θ = $\frac{\sqrt{3}}{2}$, then (cosecθ + cotθ) = ?
(a) $2\sqrt{3}$
(b) $\frac{2\sqrt{3}}{3}$
(c) $(2+\sqrt{3})$
(d) $\sqrt{3}$

Answer 1

(d) $\sqrt{3}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
sin θ = $\frac{\sqrt{3}}{2}$ = $\frac{BC}{AC}$
So, BC = $\sqrt{3}$k and AC = 2k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (2k)² $-$ ($\sqrt{3}$k)²
⇒ AB² = k²
⇒ AB = k

So, cot θ = $\frac{AB}{BC}$ = $\frac{k}{\sqrt{3}k}=\frac{1}{\sqrt{3}}$
cosec θ = $\frac{1}{\sin \theta }=\frac{2}{\sqrt{3}}$
∴ (cosec θ + cot θ) = $\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}$ = $\frac{3}{\sqrt{3}}=\sqrt{3}$