Let us first draw a right $\Delta ABC$ (see $Fig.$).

Now, we know that $\tan A =\frac{ BC }{ AB }=\frac{4}{3}$

Therefore, if $BC =4 k,$ then $AB =3 k,$ where $k$ is a positive number.

Now, by using the Pythagoras Theorem, we have

$AC ^{2}= AB ^{2}+ BC ^{2}=(4 k)^{2}+(3 k)^{2}=25 k ^{2}$

$AC =5 k$

Now, we can write all the trigonometric ratios using their definitions.

$\sin A=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

$\cos A=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$

Therefore, $\cot A=\frac{1}{\tan A}=\frac{3}{4}, \operatorname{cosec} A=\frac{1}{\sin A}=\frac{5}{4}$ and $\sec A=\frac{1}{\cos A}=\frac{5}{3}$