Question
If $3\cot\theta=4,$ find the value of $\frac{4\cos\theta-\sin\theta}{2\cos+\sin\theta}.$
✓
Answer
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, $\frac{4\cos\theta-\sin\theta}{2\cos\theta+\sin\theta}=\frac{4\times\frac{4}{5}-\frac{3}{5}}{2\times\frac{4}{5}+\frac{3}{5}}$ $=\frac{\frac{16-3}{5}}{\frac{8+3}{5}}$ $=\frac{13}{5}\times\frac{5}{11}$ $=\frac{13}{11}$
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