If 3 cot θ = 2, show that
Question:

If $3 \cot \theta=2$, show that $\left(\frac{4 \sin \theta-3 \cos \theta}{2 \sin \theta+6 \cos \theta}\right)=\frac{1}{3}$.

Solution:

It is given that $\cot \theta=\frac{2}{3}$.

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

Dividing the above expression by $\sin \theta$, we get:

$\frac{4-3 \cot \theta}{2+6 \cot \theta} \quad\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

Now, substituting the values of cot θin the above expression, we get:

$\frac{4-3\left(\frac{2}{3}\right)}{2+6\left(\frac{2}{3}\right)}$

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

i.e., LHS = RHS

Hence proved.