Question

Given : $5cosA-12sinA=0$. Evaluate : $\frac{sinA+cosA}{2cosA-sinA}$

• A. $\frac{19}{17}$
• B. $\frac{17}{19}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (B)

$\frac{17}{19}$

Here, $5cosA-12sinA=0$

$\Rightarrow tanA=\frac{5}{12}$

then, $\frac{sinA+cosA}{2cosA-sinA}$

Dividing $N^r$ and $D^r$ by cos A , we get

$\frac{tanA+1}{2-tanA}=\frac{\frac{5}{12}+1}{2-\frac{5}{12}}=\frac{17}{19}$