Trigonometric Ratios — Maths STD 10 — Question

We have,

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

In $\triangle\text{ABC},$

$\text{AC}^2=\text{AB}^2+\text{BC}^2$

$\Rightarrow\text{AC}^2=(4)^2+(3)^2$

$\Rightarrow\text{AC}^2=16+9$

$\Rightarrow\text{AC}^2=25$

$\Rightarrow\text{AC}=5$

$\therefore\ \sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{4}{5}$ and $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{3}{5}$

Now, $\frac{4\cos\theta-\sin\theta}{2\cos\theta+\sin\theta}=\frac{4\times\frac{3}{5}-\frac{4}{5}}{2\times\frac{3}{5}+\frac{4}{5}}$

$=\frac{\frac{4}{5}(3-1)}{\frac{2}{5}(3+2)}$

$=\frac{2\times2}{5}$

$=\frac{4}{5}$

Hence, $\frac{4\cos\theta-\sin\theta}{2\cos\theta+\sin\theta}=\frac{4}{5}$