Question

Given : $4\sin \Theta =3\cos \Theta$, find the value of:

$4{\cos }^2\Theta -3{\sin }^2\Theta +2$ is $3\frac{m}{25}$, m is

Answer 1

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