Question

If $5\tan \theta =4,$ then find the value of $\frac{5\sin \theta -3\cos \theta }{\sin \theta +2\cos \theta }$

Answer 1

$\tan \theta =\frac{4}{5}=\frac{oppositeside}{adjacentside}\Delta ABC$ is right angled at B,

$\angle C=\theta \Rightarrow \tan \theta =\frac{AB}{BC}=\frac{4}{5}$

If$AB=4,BC=5$

By Pythagoras theorem,

$AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{41}\therefore \sin \theta =\frac{4}{\sqrt{4}1},\cos \theta =\frac{5}{\sqrt{41}}$$\therefore \frac{5\sin \theta -3\cos \theta }{\sin \theta +2\cos \theta }=\frac{5\times \frac{4}{\sqrt{41}}-3\times \frac{5}{\sqrt{41}}}{\frac{4}{\sqrt{4}1}+2\times \frac{5}{\sqrt{4}1}}=\frac{5}{14}$