Question

If $\tan \theta =\frac{4}{3},$then find the value of $\frac{3\sin \theta +2\cos \theta }{3\sin \theta -2cos\theta }$

Answer 1

$\tan \theta =\frac{4}{3}$ (given)

$\therefore \frac{3\sin \theta +2\cos \theta }{3\sin \theta -2\cos \theta }=\frac{\frac{3\sin \theta +2\cos \theta }{\cos \theta }}{\frac{3\sin \theta -2\cos \theta }{\cos \theta }}$

$=\frac{3\tan \theta +2}{3\tan \theta -2}$

$=\frac{3\left(\frac{4}{3}\right)+2}{3\left(\frac{4}{3}\right)-2}=\frac{6}{2}=3$.