Question

If $2\tan \theta =1$, find the value of $\frac{3\cos \theta +2\sin \theta }{2\cos \theta -\sin \theta }$

• A. $\frac{8}{3}$
• B. $\frac{5}{3}$
• C. $\frac{2}{3}$
• D. $2$

Answer 1

The correct option is A $\frac{8}{3}$

$\frac{3\cos \theta +2\sin \theta }{2\cos \theta -\sin \theta }=\frac{3\frac{\cos \theta }{\cos \theta }+2\frac{\sin \theta }{\cos \theta }}{2\frac{\cos \theta }{\cos \theta }-\frac{\sin \theta }{\cos \theta }}$
[Divinding both the numerator and denominator by $\cos \theta$]
$=\frac{3+2\tan \theta }{2-\tan \theta }=\frac{3+2\times \frac{1}{2}}{2-\frac{1}{2}}$
[Given $2\tan \theta =1\therefore \tan \theta =\frac{1}{2}$]
$=\frac{3+1}{3/2}=4\times \frac{2}{3}=2\frac{2}{3}=\frac{8}{3}$