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

Given:

$3 \tan \theta=4$

Therefore,

$\tan \theta=\frac{4}{3}$...(1)

Now, we know that $\tan \theta=\frac{\sin \theta}{\cos \theta}$

Therefore equation (1) becomes

$\frac{\sin \theta}{\cos \theta}=\frac{4}{3}$....(2)

Now, by applying Invertendo to equation (2)

We get,

$\frac{\cos \theta}{\sin \theta}=\frac{3}{4}$

Now, multiplying by 4 on both sides

We get,

$4 \times \frac{\cos \theta}{\sin \theta}=4 \times \frac{3}{4}$

Therefore

$\frac{4 \cos \theta}{\sin \theta}=\frac{3}{1}$

Now by applying dividendo in above equation

We get,

$\frac{4 \cos \theta-\sin \theta}{\sin \theta}=\frac{3-1}{1}$

$\frac{4 \cos \theta-\sin \theta}{\sin \theta}=\frac{2}{1}$...(4)

Now, multiplying by 2 on both sides of equation (3)

We get,

$2 \times \frac{\cos \theta}{\sin \theta}=2 \times \frac{3}{4}$

Therefore

$\frac{2 \cos \theta}{\sin \theta}=\frac{3}{2}$

Now by applying componendo in above equation

We get,

$\frac{2 \cos \theta+\sin \theta}{\sin \theta}=\frac{3+2}{2}$

$\frac{2 \cos \theta+\sin \theta}{\sin \theta}=\frac{5}{2}$...(5)

Now, by dividing equation (4) by equation (5)

We get,

$\frac{\frac{4 \cos \theta-\sin \theta}{\sin \theta}}{\frac{2 \cos \theta+\sin \theta}{\sin \theta}}=\frac{\frac{2}{1}}{\frac{5}{2}}$

Therefore,

$\frac{4 \cos \theta-\sin \theta}{\sin \theta} \times \frac{\sin \theta}{2 \cos \theta+\sin \theta}=\frac{2}{1} \times \frac{2}{5}$

Therefore, on L.H.S $\sin \theta$ cancels and we get,

$\frac{4 \cos \theta-\sin \theta}{2 \cos \theta+\sin \theta}=\frac{2}{1} \times \frac{2}{5}$

Therefore,

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

Hence,

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