Question

If $\theta$ lies in the first quadrant and $5 \tan \theta=4$, then $\frac{5 \sin \theta-3 \cos \theta}{\sin \theta+2 \cos \theta}$ is equal to

• A. $\frac{5}{14}$
• B. $\frac{3}{14}$
• C. $\frac{1}{14}$
• D. 0

Answer 1

Answer: (A)

Given, $\tan \theta=\frac{4}{5}$
$\therefore \sin \theta=\frac{4}{\sqrt{41}}$ and $\cos \theta=\frac{5}{\sqrt{41}}$
Now, $\frac{5 \sin \theta-3 \cos \theta}{\sin \theta+2 \cos \theta}=\frac{5 \times \frac{4}{\sqrt{41}}-3 \times \frac{5}{\sqrt{41}}}{\frac{4}{\sqrt{41}}+2 \times \frac{5}{\sqrt{41}}}=\frac{5}{14}$