Question

If $6\tan A-5=0$, find the value of: $\frac{3\sin A-\cos A}{6\cos A+\sin A}$ is $\frac{9}{m}$, then m is

Answer 1

$6\tan A-5=0$
$\tan A=\frac{5}{6}$
Now, $\frac{3\sin A-\cos A}{6\cos A+\sin A}$
Multiply and divide by $\cos A$
= $\frac{3\tan A-1}{6+\tan A}$
= $\frac{3\left(\frac{5}{6}\right)-1}{6+\frac{5}{6}}$
= $\frac{\frac{5}{2}-1}{\frac{41}{6}}$
= $\frac{\frac{5-2}{2}}{\frac{41}{6}}$
= $\frac{9}{41}$