If 3 cot θ = 4, find the value of 4 cos θ−sin θ2 cos θ+sin θ.

Solution:

We have:

$3 \cot \theta=4$

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

Since we know that in right angle triangle

$\cot \theta=\frac{\text { Base }}{\text { perpendicular }}$

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$

Hypotenuse $=\sqrt{(3)^{2}+(4)^{2}}$

Hypotenuse $=\sqrt{25}$

Hypotenuse $=5$

Now, we find $\frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}$

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

$\Rightarrow \frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}=\frac{\frac{16}{5}-\frac{3}{5}}{\frac{8}{5}+\frac{3}{5}}$

$\Rightarrow \frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}=\frac{\frac{13}{5}}{\frac{11}{5}}$

$\Rightarrow \frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}=\frac{13}{11}$

Hence the value of $\frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}$ is $\frac{13}{11}$