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Question Number 27429 by Tinkutara last updated on 06/Jan/18

Answered by mrW1 last updated on 07/Jan/18

$${let}^{\prime }sseepoint2asexample,$$$$\frac{v_2\cos {\theta }_2}{v_1\cos {\theta }_1}=e$$$$\Rightarrow v_2\cos {\theta }_2=e{v}_1\cos {\theta }_1...\left(i\right)$$$${mv}_2\sin {\theta }_2={mv}_1\sin {\theta }_1$$$$\Rightarrow v_2\sin {\theta }_2=v_1\sin {\theta }_1...\left(ii\right)$$$$\left(ii\right)/\left(i\right):$$$$\Rightarrow \tan {\theta }_2=\frac{1}{e}\tan {\theta }_1$$$$$$$$similarlyatpoint1:$$$$\Rightarrow \tan {\theta }_1=\frac{1}{e}\tan \theta$$$$$$$$\Rightarrow \tan {\theta }_2=\frac{1}{e}\tan {\theta }_1=\frac{1}{e^2}\tan \theta$$$$$$$$2({\theta }_2+{\theta }_1+\theta )=\pi$$$${\theta }_2+{\theta }_1=\frac{\pi }{2}-\theta$$$$\tan ({\theta }_2+{\theta }_1)=\tan (\frac{\pi }{2}-\theta )$$$$\frac{\tan {\theta }_2+\tan {\theta }_1}{1-\tan {\theta }_1\tan {\theta }_2}=\frac{1}{\tan \theta }$$$$\frac{\frac{\tan \theta }{e^2}+\frac{\tan \theta }{e}}{1-\frac{{\tan }^2\theta }{e^3}}=\frac{1}{\tan \theta }$$$$\frac{e(e+1)\tan \theta }{e^3-{\tan }^2\theta }=\frac{1}{\tan \theta }$$$$e(e+1){\tan }^2\theta =e^3-{\tan }^2\theta$$$$(e^2+e+1){\tan }^2\theta =e^3$$$$\Rightarrow {\tan }^2\theta =\frac{e^3}{e^2+e+1}$$$$$$$$\Rightarrow Answer\left(4\right)$$

Commented by mrW1 last updated on 07/Jan/18

Commented by Tinkutara last updated on 08/Jan/18

Thank you very much Sir! I got the answer.

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