Question

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

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{1}{6}$
• D. $\frac{5}{6}$

Answer 1

Answer: (C)

$\frac{1}{6}$

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

Now, $\frac{4\sin \theta -2\cos \theta }{4\sin \theta +3\cos \theta }$

Dividing both numerator and denominator by $\cos \theta$ we get

$=\frac{\frac{4\sin \theta }{\cos \theta }-2\frac{\cos \theta }{\cos \theta }}{\frac{4\sin \theta }{\cos \theta }+3\frac{\cos \theta }{\cos \theta }}$

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

$=\frac{4\times \frac{3}{4}-2}{4\times \frac{3}{4}+3}[\because 3\tan \theta =\frac{3}{4}]$

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

$=\frac{1}{6}$

$\therefore \frac{4\sin \theta -2\cos \theta }{4\sin \theta +3\cos \theta }=\frac{1}{6}$