Question

If $3tan\theta$ = 4 then $\frac{4cos\theta -sin\theta }{3cos\theta +sin\theta }$

Answer 1

given, $3\tan \theta =4$

$\Rightarrow \frac{3\sin \theta }{\cos \theta }=4$

$\Rightarrow 3\sin \theta =4\cos \theta$ ..........$\left(1\right)$

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

$=\frac{3\sin \theta -\sin \theta }{\frac{3\sin \theta }{4}+\sin \theta }$ from $\left(1\right)$

$=\frac{8\sin \theta }{3\sin \theta +4\sin \theta }$

$=\frac{8\sin \theta }{7\sin \theta }$

$=\frac{8}{7}$