Question
The equations of $L_{1}$ and $L_{2}$ are $y = mx$ and $y = nx$, respectively. Suppose $L_{1}$ make twice as large of an angle with the horizontal (measured counterclockwise from the positive $x$-axis) as does $L_{2}$ and that $L_{1}$ has 4 times the slope of $L_{2}$. If $L_{1}$ is not horizontal, then the value of the product $(mn)$ equals:
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Solution
Sol. [Key Concept: $L_{2}$ makes twice the angle and $4 \times$ the slope of $L_{1}$ ⇒ $\tan\ 2\theta = 4\tan\ \theta$ gives the product $mn$.]
Let $m = tan2\theta$ and $n = tan\theta$ and $m = 4n\ $ and $\ m \neq 0$,
$$\therefore\ tan2\theta \neq 0;\ \therefore\ \theta \neq 0$$
$$\begin{matrix} & tan2\theta = 4tan\theta \\ & \frac{2tan\theta}{1 - \tan^{2}\theta} = 4tan\theta \\ \therefore\ & 1 = 2 - 2\tan^{2}\theta \end{matrix}$$
$$\begin{matrix} 2\tan^{2}\theta = 1 \Rightarrow \tan^{2}\theta = 1/2 \\ \therefore\ mn = tan2\theta \cdot tan\theta = \frac{2\tan^{2}\theta}{1 - \tan^{2}\theta} \\ = \frac{1}{1 - (1/2)} = 2 \end{matrix}$$
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