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Time\ of\ flight\ T1\ = \frac{ 2\ u\ sin\ \alpha}{ g} \\ \\ \ \ \ T2 = \frac{2\ u\ sin\ (90 - \alpha)} { g } = \frac{2 u cos\ \alpha}{ g} \\ \\ \frac{T1}{T2} = tan\ \alpha \\ \\

the answer is this
i am sorry kvn murthy it was a mistake by me
this is the correct answer
i thought of other answer
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