If 4tan θ = 3 then prove that

Solution:

Given: $4 \tan \theta=3$

$\Rightarrow \tan \theta=\frac{3}{4}$

Since, $\tan \theta=\frac{P}{B}$

$\Rightarrow P=3$ and $B=4$

Using Pythagoras theorem,

$P^{2}+B^{2}=H^{2}$

$\Rightarrow 3^{2}+4^{2}=H^{2}$

$\Rightarrow H^{2}=9+16=25$

$\Rightarrow H=5$

Therefore,

$\sin \theta=\frac{P}{H}=\frac{3}{5}$

$\cos \theta=\frac{B}{H}=\frac{4}{5}$

$\sin \theta \times \cos \theta=\frac{3}{5} \times \frac{4}{5}$

$=\frac{12}{25}$

Hence, $\sin \theta \times \cos \theta=\frac{12}{25}$.