Question

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

Answer 1

$6\tan A-5=0$
$\tan A=\frac{5}{6}$
Now, $+\frac{3sinA-cosA}{5cosA+9sinA}$
Multiply and Divide by $\cos A$
= $\frac{3tanA-1}{5+9tanA}$
= $\frac{3\left(\frac{5}{6}\right),-1}{5+9\left(\frac{5}{6}\right)}$
= $\frac{\frac{5}{2},-1}{5+\frac{15}{2}}$
= $\frac{5-2}{10+15}$
= $\frac{3}{25}$