If cot θ = 2, find the value of all T-ratios of θ.

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

Now, we know that $\cot \theta=\frac{\text { base }}{\text { Perpendicular }}=\frac{B C}{A B}=2$.

So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2

$\Rightarrow \mathrm{AC}=\sqrt{5} \mathrm{k}$

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

$\sin \theta=\frac{A B}{A C}=\frac{k}{\sqrt{5} k}=\frac{1}{\sqrt{5}}$

$\cos \theta=\frac{B C}{A C}=\frac{2 k}{\sqrt{5} k}=\frac{2}{\sqrt{5}}$

$\therefore \tan \theta=\frac{1}{\cot \theta}=\frac{1}{2}, \operatorname{cosec} \theta=\frac{1}{\sin \theta}=\sqrt{5}$ and $\sec \theta=\frac{1}{\cos \theta}=\frac{\sqrt{5}}{2}$