Question

Solve the following with the help of relations between trignometric ratios
If $\tan \theta =\frac{4}{3}$, then find the value of $\frac{3\sin \theta +2\cos \theta }{3\sin \theta -2\cos \theta }$

Answer 1

$\tan \theta =\frac{4}{3}$

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

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

$\frac{3\times \frac{4}{3}+2}{3\times \frac{4}{3}-2}$

$=\frac{4+2}{4-2}=\frac{6}{2}=3.$

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