# If 3 cot θ = 2, find the value of 4 sin θ−3 cos θ2 sin θ+6 cos θ.

Question:

If $3 \cot \theta=2$, find the value of $\frac{4 \sin \theta-3 \cos \theta}{2 \sin \theta+6 \cos \theta}$.

Solution:

Given:

$3 \cot \theta=2$

Therefore,

$\cot \theta=\frac{2}{3}$..(1)

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

Therefore equation (1) becomes

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

Now, by applying Invertendo to equation (2)

We get,

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

Now, multiplying by $\frac{4}{3}$ on both sides

We get,

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

Therefore, 3 cancels out on R.H.S and

We get,

$\frac{4 \sin \theta}{3 \cos \theta}=\frac{2}{1}$

Now by applying dividendo in above equation

We get,

$\frac{4 \sin \theta-3 \cos \theta}{3 \sin \theta}=\frac{2-1}{1}$

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

Now, multiplying by $\frac{2}{6}$ on both sides of equation (3)

We get,

$\frac{2}{6} \times \frac{\sin \theta}{\cos \theta}=\frac{2}{6} \times \frac{3}{2}$

Therefore, 2 cancels out on R.H.S and

We get,

$\frac{2 \sin \theta}{6 \cos \theta}=\frac{3}{6}$

$\frac{2 \sin \theta}{6 \cos \theta}=\frac{1}{2}$

Now by applying componendo in above equation

We get,

$\frac{2 \cos \theta+6 \sin \theta}{6 \sin \theta}=\frac{1+2}{2}$

$\frac{2 \cos \theta+6 \sin \theta}{6 \sin \theta}=\frac{3}{2}$....(5)

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

We get,

$\frac{\frac{4 \sin \theta-3 \cos \theta}{3 \sin \theta}}{\frac{2 \cos \theta+6 \sin \theta}{6 \sin \theta}}=\frac{\frac{1}{1}}{\frac{3}{2}}$

Therefore,

$\frac{4 \sin \theta-3 \cos \theta}{3 \sin \theta} \times \frac{6 \sin \theta}{2 \cos \theta+6 \sin \theta}=\frac{1}{1} \times \frac{2}{3}$

$\frac{4 \sin \theta-3 \cos \theta}{3 \sin \theta} \times \frac{2 \times(3 \sin \theta)}{2 \cos \theta+6 \sin \theta}=\frac{1}{1} \times \frac{2}{3}$

Therefore, on L.H.S $(3 \sin \theta)$ cancels out and we get,

$\frac{2 \times(4 \sin \theta-3 \cos \theta)}{2 \cos \theta+6 \sin \theta}=\frac{2}{3}$

Now, by taking 2 in the numerator of L.H.S on the R.H.S

We get,

$\frac{4 \sin \theta-3 \cos \theta}{2 \cos \theta+6 \sin \theta}=\frac{2}{3 \times 2}$

Hence,

$\frac{4 \sin \theta-3 \cos \theta}{2 \cos \theta+6 \sin \theta}=\frac{1}{3}$