Given $\tan A = \frac{4}{3},$ find the other trigonometric ratios of the $\angle A$.

(N/A) Let us first draw a right $\Delta ABC$ where $\angle B = 90^{\circ}$.
Now,we know that $\tan A = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{BC}{AB} = \frac{4}{3}$.
Therefore,if $BC = 4k,$ then $AB = 3k,$ where $k$ is a positive number.
Now,by using the Pythagoras Theorem,we have:
$AC^2 = AB^2 + BC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$.
$AC = 5k$.
Now,we can write all the trigonometric ratios using their definitions:
$\sin A = \frac{BC}{AC} = \frac{4k}{5k} = \frac{4}{5}$.
$\cos A = \frac{AB}{AC} = \frac{3k}{5k} = \frac{3}{5}$.
$\cot A = \frac{1}{\tan A} = \frac{3}{4}$.
$\operatorname{cosec} A = \frac{1}{\sin A} = \frac{5}{4}$.
$\sec A = \frac{1}{\cos A} = \frac{5}{3}$.