Question
If $3\tan\theta=4,$ show that $\frac{(4\cos\theta-\sin\theta)}{(2\cos\theta+\sin\theta)}=\frac45.$
✓
Answer
$3\tan\theta=4\Rightarrow\tan\theta=\frac43$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac43$
Let BC = 4 and AB = 3
Then, by pythagoras theoram,
$AC^2 = AB^2 + BC^2$
$= 3^2 + 4^2 = 9 + 16 = 25$
$\Rightarrow AC = 5$
Now,
$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac45$
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac35$
$\therefore\text{L.H.S.}=\frac{(4\cos\theta-\sin\theta)}{(2\cos\theta+\sin\theta)}$
$=\frac{4\times\frac35-\frac45}{2\times\frac{3}{5}+\frac{4}{5}}$
$=\frac{\frac85}{\frac{10}{5}}$
$=\frac{8}{5}\times\frac12$
$=\frac45$
$=\text{R.H.S.}$
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