Question

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

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

Answer 1

Answer: (C)

$\frac{1}{3}$

Given, $\frac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }$
[Dividing both numerator anddenominator by $\sin \theta$]
$=\frac{4\frac{\sin \theta }{\sin \theta }-\frac{3\cos \theta }{\sin \theta }}{4\frac{\sin \theta }{\sin \theta }+\frac{6\cos \theta }{\sin \theta }}$
$=\frac{4-3\cot \theta }{2+6\cot \theta }=\frac{4-3\times \frac{2}{3}}{2+6\times \frac{2}{3}}[\because 3\cot \theta =2]$
$=\frac{4-2}{2+4}=\frac{2}{6}=\frac{1}{3}[\because \cot \theta =\frac{2}{3}]$