Question

If $3\cos A=4\sin A;$ find the value of: $4{\cos }^{2}A-3{\sin }^{2}A+2$ is $3\frac{m}{25}$, m is

Answer 1

$3\cos A=4\sin A$
$\tan A=\frac{3}{4}$
$\tan A=\frac{P}{B}=\frac{3}{4}$
Now, Using Pythagoras Theorem,
$H^2=P^2+B^2$
$H^2=3^2+4^2$
$H=5$
Now, $4co{s}^{2}A-3si{n}^{2}A+2$
= $4(\frac{B}{H}{)}^2-3(\frac{P}{H}{)}^2+2$
= $4(\frac{4}{5}{)}^2-3(\frac{3}{5}{)}^2+2$
= $\frac{64}{25}-\frac{27}{25}+2$
= $\frac{64-27+50}{25}$
= $\frac{87}{25}$
= $3\frac{12}{25}$