Prove the following

Given,

$\tan A=\frac{3}{4}=\frac{P}{B}=\frac{\text { Perpendicular }}{\text { Base }}$

Let $P=3 k$ and $B=4 k$

By Pythagoras theorem,

$H^{2}=P^{2}+B^{2}=(3 k)^{2}+(4 k)^{2}$

$=9 k^{2}+16 k^{2}=25 k^{2}$

$\begin{array}{lll}\Rightarrow & H=5 k & \text { [since, side cannot be negative] }\end{array}$

$\therefore \quad \sin A=\frac{P}{H}=\frac{3 k}{5 k}=\frac{3}{5}$ and $\cos A=\frac{B}{H}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, $\quad \sin A \cos A=\frac{3}{5} \cdot \frac{4}{5}=\frac{12}{25}$

Hence proved.