Question

If $sinA:cosA=4:7$, then what is the value of $\frac{7sinA-3cosA}{7sinA+2cosA}$?

• A. $\frac{1}{6}$
• B. $\frac{2}{12}$
• C. $\frac{2}{6}$
• D. $\frac{3}{9}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{6}$
B $\frac{2}{12}$

$sinA:cosA=4:7$

$\frac{sinA}{cosA}=\frac{4}{7}$

$\frac{7sinA-3cosA}{7sinA+2cosA}$

$=\frac{7\frac{sinA}{cosA}-3}{7\frac{sinA}{cosA}+2}$

$=\frac{7\left(\frac{4}{7}\right)-3}{7\left(\frac{4}{7}\right)+2}$

$=\frac{4-3}{4+2}$

$=\frac{1}{6}$