Question

If 6 tan a-5=0,find the value of:
3 sin a-cos a/5 cos a+9 sin a

Answer 1

$\huge{ \boxed{ \boxed{ \bold \red{Given:}}}}$

$6\;tanA-5=0$

$\implies\,tanA=\frac{5}{6}$

$\implies\frac{sinA}{cosA}=\frac{5}{6}$

$\text{Then, }sinA=5k\;\;\&\;\;cosA=6k$

$\displaystyle\frac{3\,sinA-cosA}{5\,cosA+9\,sinA}$

$=\displaystyle\frac{3(5k)-6k}{5(6k)+9(5k)}$

$=\displaystyle\frac{15k-6k}{30k+45k}$

$=\displaystyle\frac{9k}{75k}$

$=\displaystyle\frac{3}{25}$

$\implies\boxed{\bf\frac{3\,sinA-cosA}{5\,cosA+9\,sinA}=\frac{3}{25}}$