If $A$ lies in the first quadrant and $6 \tan A=5$, then the value of $\frac{8 \sin A-4 \cos A}{\cos A+2 \sin A}$ is:

1 -2 4 16

1

6tanA = 5 tanA = \(\frac{5}{6}\) { we know, tanA = \(\frac{P}{B}\) } Now, \(\frac{8sinA - 4cosA}{ cosA + 2sinA }\) = \(\frac{8 × P/H - 4×B/H}{ B/H + 2×P/H }\) = \(\frac{8 × P - 4×B}{ B + 2× P }\) = \(\frac{8 × 5 - 4×6}{ 6+ 2× 5 }\) = \(\frac{16}{ 16 }\) = 1