Prove that

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

Now, we know that $\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}=\frac{A C}{A B}=\frac{\sqrt{10}}{1}$.

So, if $A C=(\sqrt{10}) k$, then $A B=k$, where $k$ is a positive number.

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

$\Rightarrow B C^{2}=A C^{2}-A B^{2}=10 k^{2}-k^{2}$

$\Rightarrow B C^{2}=9 k^{2}$

$\Rightarrow \mathrm{BC}=3 \mathrm{k}$

Now, finding the other T-ratios using their definitions, we get:

$\tan \theta=\frac{A B}{B C}=\frac{k}{3 k}=\frac{1}{3}$

$\cos \theta=\frac{B C}{A C}=\frac{3 k}{\sqrt{10} k}=\frac{3}{\sqrt{10}}$

$\therefore \sin \theta=\frac{1}{\operatorname{cosec} \theta}=\frac{1}{\sqrt{10}}, \cot \theta=\frac{1}{\tan \theta}=3$ and $\sec \theta=\frac{1}{\cos \theta}=\frac{\sqrt{10}}{3}$