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A vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles substended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $(\tan \alpha, \tan \beta)$ is equal to
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The correct answer is:
$\left(\frac{2}{9}, \frac{1}{4}\right)$
Let $A C$ be a pole and point $P$ be the position on of the ground.
Given,
$\theta=\tan ^{-1} \frac{1}{2}$
$\Rightarrow \quad \tan \theta=\frac{1}{2}$
Also, $\quad \theta=\alpha+\beta$
$\begin{aligned} \Rightarrow & \tan \theta & =\tan (\alpha+\beta) \\ \Rightarrow & \frac{1}{2} & =\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\end{aligned}$
(a) When $(\tan \alpha, \tan \beta)=\left(\frac{1}{4}, \frac{1}{5}\right)$
$\therefore$ RHS $=\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4} \times \frac{1}{5}}=\frac{\frac{9}{20}}{\frac{19}{20}}$
$=\frac{9}{19} \neq \frac{1}{2}$, not true
(b)
$\begin{aligned} & \text { When }(\tan \alpha, \tan \beta)=\left(\frac{1}{5}, \frac{2}{9}\right) \\ & \begin{aligned} \text { RHS } & =\frac{\frac{1}{5}+\frac{2}{9}}{1-\frac{1}{5} \times \frac{2}{9}}=\frac{\frac{19}{45}}{\frac{43}{45}} \\ = & \frac{19}{43} \neq \frac{1}{2}, \text { not true }\end{aligned}\end{aligned}$
(c)
When $(\tan \alpha, \tan \beta)=\left(\frac{2}{9}, \frac{1}{4}\right)$ RHS $=\frac{\frac{2}{9}+\frac{1}{4}}{1-\frac{2}{9} \times \frac{1}{4}}=\frac{\frac{17}{36}}{\frac{34}{36}}=\frac{1}{2}$, true
Given,
$\theta=\tan ^{-1} \frac{1}{2}$
$\Rightarrow \quad \tan \theta=\frac{1}{2}$
Also, $\quad \theta=\alpha+\beta$
$\begin{aligned} \Rightarrow & \tan \theta & =\tan (\alpha+\beta) \\ \Rightarrow & \frac{1}{2} & =\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\end{aligned}$
(a) When $(\tan \alpha, \tan \beta)=\left(\frac{1}{4}, \frac{1}{5}\right)$
$\therefore$ RHS $=\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4} \times \frac{1}{5}}=\frac{\frac{9}{20}}{\frac{19}{20}}$
$=\frac{9}{19} \neq \frac{1}{2}$, not true
(b)
$\begin{aligned} & \text { When }(\tan \alpha, \tan \beta)=\left(\frac{1}{5}, \frac{2}{9}\right) \\ & \begin{aligned} \text { RHS } & =\frac{\frac{1}{5}+\frac{2}{9}}{1-\frac{1}{5} \times \frac{2}{9}}=\frac{\frac{19}{45}}{\frac{43}{45}} \\ = & \frac{19}{43} \neq \frac{1}{2}, \text { not true }\end{aligned}\end{aligned}$
(c)
When $(\tan \alpha, \tan \beta)=\left(\frac{2}{9}, \frac{1}{4}\right)$ RHS $=\frac{\frac{2}{9}+\frac{1}{4}}{1-\frac{2}{9} \times \frac{1}{4}}=\frac{\frac{17}{36}}{\frac{34}{36}}=\frac{1}{2}$, true
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