Prove that:
If $\tan A = \frac{3}{4},$ then $\sin A \cos A = \frac{12}{25}$

(N/A) Given,$\tan A = \frac{3}{4} = \frac{P}{B} = \frac{\text{Perpendicular}}{\text{Base}}$.
Let $P = 3k$ and $B = 4k$.
By Pythagoras theorem,
$H^2 = P^2 + B^2 = (3k)^2 + (4k)^2$
$H^2 = 9k^2 + 16k^2 = 25k^2$
$\Rightarrow H = 5k$ [since,side length cannot be negative].
Now,$\sin A = \frac{P}{H} = \frac{3k}{5k} = \frac{3}{5}$ and $\cos A = \frac{B}{H} = \frac{4k}{5k} = \frac{4}{5}$.
Therefore,$\sin A \cos A = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$.
Hence proved.