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