Question:
If $\sin \theta=\frac{\sqrt{3}}{2}$ then $(\operatorname{cosec} \theta+\cot \theta)=?$
(a) $(2+\sqrt{3})$
(b) $2 \sqrt{3}$
(c) $\sqrt{2}$
(d) $\sqrt{3}$
Solution:
Given : $\sin \theta=\frac{\sqrt{3}}{2}$
Since, $\sin \theta=\frac{P}{H}$
$\Rightarrow P=\sqrt{3}$ and $H=2$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow(\sqrt{3})^{2}+B^{2}=2^{2}$
$\Rightarrow B^{2}=4-3$
$\Rightarrow B^{2}=1$
$\Rightarrow B=1$
Therefore,
$\operatorname{cosec} \theta=\frac{H}{P}=\frac{2}{\sqrt{3}}$
$\cot \theta=\frac{B}{P}=\frac{1}{\sqrt{3}}$
Now,
$(\operatorname{cosec} \theta+\cot \theta)=\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)$
$=\left(\frac{2+1}{\sqrt{3}}\right)$
$=\left(\frac{3}{\sqrt{3}}\right)$
$=\sqrt{3}$
Hence, the correct option is (d).