Question

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

Answer 1

Given,

$3\tan \theta =4$

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

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

$=\frac{4cos\theta -sin\theta }{2cos\theta +sin\theta }\times \frac{\cos \theta }{\cos \theta }$ ----------( Multiply and divide by $\cos \theta$ )

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

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

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

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

$=\frac{12-4}{6+4}$

$=\frac{8}{10}$

$=\frac{4}{5}$