Question

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

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

Answer 1

Answer: (C)

$\frac{4}{5}$

Given,$3\tan \alpha =4\Rightarrow \tan \alpha =\frac{4}{3}$
$\frac{4\cos \alpha -\sin \alpha }{2\cos \alpha +\sin \alpha }=\frac{4-\tan \alpha }{2+\tan \alpha }$

($\text{On dividing numerator and denominator by}\cos \alpha$)

$=\frac{4-\frac{4}{3}}{2+\frac{4}{3}}$ $=\frac{12-4}{6+4}$ $=\frac{8}{10}$ $=\frac{4}{5}$

Option C is correct.