Question

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

Answer 1

given $3\tan \theta =4⟹\tan \theta =\frac{4}{3}$

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