Question

Given that 4tanA = 3 , then the value of$\frac{4sinA-cosA}{4sinA+cosA}$ is

• A. 1/2
• B. 1/√2
• C. 0
• D. 1/√2

Answer 1

Answer:

A. 1/2

4 tanA = 3

tanA = 3/4

tan A = opposite side/adjascent side

hypotnuse² = 3² + 4²

hypotnuse² = 9 + 16

hypotnuse = 5

sinA = 3/5

cosA= 4/5

$\frac{4sinA-cosA}{4sinA+cosA} =\frac{(4 \times \frac{3}{5})-\frac{4}{5}}{(4 \times \frac{3}{5})+\frac{4}{5}}$

$=\frac{\frac{12}{5}-\frac{4}{5}}{\frac{12}{5}+\frac{4}{5}}$

$\frac{8/5}{16/5}=\frac{8}{16}=\frac{1}{2}$