Question
If $3\cot\theta=2,$ show that $\frac{(4\sin\theta-3\cos\theta)}{(2\sin\theta+6\cos\theta)}=\frac13.$
✓
Answer
Given:
$\cot\theta=2,\ \therefore\cot\theta=\frac23=\frac{2\text{k}}{3\text{k}}$
Let us draw a $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theoram, we have
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$=\big(2\text{k}\big)^2+\big(3\text{k}\big)^2$
$=4\text{k}^2+9\text{k}^2=13\text{k}^2$
$\Rightarrow\text{AC}=\sqrt{13}\text{k}$
$\therefore\sin\theta=\frac{3\text{k}}{\sqrt{13\text{k}}}=\frac{3}{\sqrt{13}}$
$\cos\theta=\frac{2\text{k}}{\sqrt{13}\text{k}}=\frac{2}{\sqrt{13}}$
$\therefore\text{L.H.S.}=\frac{(4\sin\theta-3\cos\theta)}{(2\sin\theta+6\cos\theta)}$
$=\frac{4\times\frac{3}{\sqrt{13}}-3\times\frac{2}{\sqrt{13}}}{2\times\frac{3}{\sqrt{13}}+6\times\frac{2}{\sqrt{13}}}$
$=\frac{\frac{12-6}{\sqrt{13}}}{\frac{6+12}{\sqrt{13}}}$
$=\frac{6}{18}=\frac13$
$=\text{R.H.S.}$
$\cot\theta=2,\ \therefore\cot\theta=\frac23=\frac{2\text{k}}{3\text{k}}$
Let us draw a $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theoram, we have
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$=\big(2\text{k}\big)^2+\big(3\text{k}\big)^2$
$=4\text{k}^2+9\text{k}^2=13\text{k}^2$
$\Rightarrow\text{AC}=\sqrt{13}\text{k}$
$\therefore\sin\theta=\frac{3\text{k}}{\sqrt{13\text{k}}}=\frac{3}{\sqrt{13}}$
$\cos\theta=\frac{2\text{k}}{\sqrt{13}\text{k}}=\frac{2}{\sqrt{13}}$
$\therefore\text{L.H.S.}=\frac{(4\sin\theta-3\cos\theta)}{(2\sin\theta+6\cos\theta)}$
$=\frac{4\times\frac{3}{\sqrt{13}}-3\times\frac{2}{\sqrt{13}}}{2\times\frac{3}{\sqrt{13}}+6\times\frac{2}{\sqrt{13}}}$
$=\frac{\frac{12-6}{\sqrt{13}}}{\frac{6+12}{\sqrt{13}}}$
$=\frac{6}{18}=\frac13$
$=\text{R.H.S.}$
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