Question

If $3\cos \theta =2\sin \theta$, then the value of $\frac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }$ is:

• A. $\frac{1}{8}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{3}$

Given that:

$3\cos \theta =2\sin \theta$

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

$\cot \theta =\frac{2}{3}$

Now, $\frac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }=\frac{\sin \theta [4-3\frac{\cos \theta }{\sin \theta }]}{\sin \theta [2+6\frac{\cos \theta }{\sin \theta }]}$

Divided and multiplied the numerator and denominator by $\sin \theta$.

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

$=\frac{4-(3\times \frac{2}{3})}{2+(6\times \frac{2}{3})}$

$=\frac{4-2}{2+4}$

$=\frac{2}{6}$

$=\frac{1}{3}.$

Hence, the answer is $=\frac{1}{3}.$