A cylindrical hall has a horizontal smooth floor. A ball is projected along the floor from A point on the wall in a direction making an angle $θ$ with the radius through the point the ball returns back to the initial point after two impacts with the wall. If the coefficient of restitution is e then tan² $θ$ will be

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$\frac{1 + e + e^{2}}{e^{3}}$

$\frac{1 + e}{e^{2}}$

$\frac{e^{2}}{1 + e}$

$\frac{e^{3}}{1 + e + e^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} \tan α = \frac{usin θ}{eucos θ} = \frac{\tan θ}{e} \\ ∴ \tan α = \frac{1}{e} \tan θ \\ ∴ \tan β = \frac{1}{e} \tan θ \\ 2 θ + 2 α + 2 β = 180^{\circ} \\ α + β = (90 - θ) \\ \tan (α + β) = \cot θ \\ \Rightarrow \frac{\tan α + \tan β}{1 - \tan αtan β} = \frac{1}{\tan θ} \end{array}$
$\begin{array}{l} \Rightarrow \frac{\frac{1}{\tan θ + \frac{1}{e^{2}} \tan θ}}{1 - \frac{1}{e^{3}} \tan^{2} θ} = \frac{1}{\tan θ} \\ \Rightarrow \tan^{2} θ (\frac{1}{e} + \frac{1}{e^{2}}) = 1 - \frac{1}{e^{3}} \tan^{2} θ \\ \Rightarrow \tan^{2} θ (\frac{1}{e} + \frac{1}{e^{2}} + \frac{1}{e^{3}}) = 1 \\ \tan^{2} θ (\frac{1}{\frac{1}{e} + \frac{1}{e^{2}} + \frac{1}{e^{3}}}) = \frac{e^{3}}{1 + e + e^{2}} \end{array}$