Solve this

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).