Question

If 4 tan θ = 3, then $\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }$ is equal to _________.

Answer 1

$\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }$
$=\frac{{\displaystyle \frac{4\sin \theta }{\cos \theta }}-1}{{\displaystyle \frac{4\sin \theta }{\cos \theta }}+1}$ (Dividing numerator and denominator by cosθ)
$=\frac{4\tan \theta -1}{4\tan \theta +1}$
$$\begin{gathered} =\frac{3-1}{3+1}\left(\because 4\tan \theta =3\right) \\ =\frac{2}{4} \\ =\frac{1}{2} \end{gathered}$$

If 4 tan θ = 3, then $\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }$ is equal to $\underline{\frac{1}{2}}$.