Solve this

Let us first draw a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$.

Now, we know that $\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{\sqrt{3}}{2}$.

So, if $\mathrm{AB}=\sqrt{3} k$, then $\mathrm{AC}=2 k$, where $k$ is a positive number.

Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 

$\Rightarrow \mathrm{BC}^{2}=\mathrm{AC}^{2}-\mathrm{AB}^{2}=(2 k)^{2}-(\sqrt{3} k)^{2}$

⇒ BC2 = 4k2 -">−- 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:

$\cos \theta=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}$

$\tan \theta=\frac{A B}{B C}=\frac{\sqrt{3} k}{k}=\sqrt{3}$

$\therefore \cot \theta=\frac{1}{\tan \theta}=\frac{1}{\sqrt{3}}, \operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{2}{\sqrt{3}}$ and $\sec \theta=\frac{1}{\cos \theta}=2$