Q. 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
Explanation
(a)
$\tan{\theta = \frac{4}{5}}$
$\therefore\ \sin{\theta = \frac{4}{\sqrt{41}}},\ \cos{\theta = \frac{5}{\sqrt{41}}}$
Now, $\frac{5\sin{\theta - 3\cos\theta}}{\sin{\theta + 2\cos\theta}} = \frac{5 \times \frac{4}{\sqrt{41}}\ - \ \frac{3 \times 5}{\sqrt{41}}}{\frac{4}{\sqrt{41}}\ + 2 \times \frac{5}{\sqrt{41}}} = \frac{5}{14}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.