Trigonometric Ratios — Maths STD 10 — Question

We have,

$3\cot\theta=4$

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

In $\triangle\text{ABC},$

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

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

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

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

$\Rightarrow\text{AC}=5$

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

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

$=\frac{\frac{16-3}{5}}{\frac{8+3}{5}}$

$=\frac{13}{5}\times\frac{5}{11}$

$=\frac{13}{11}$