JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    If \[\sin (\alpha +\beta )=1,\,\,\sin (\alpha -\beta )=\frac{1}{2}\] then \[\tan \,(a+2\beta )\tan \,(2\alpha +\beta )\] is equal to   AIEEE  Solved  Paper-2002

    A) 1   

    B) - 1             

    C) zero                         

    D) None of these

    Correct Answer: A

    Solution :

      \[\sin \frac{\pi }{2}=1\] and \[\sin \frac{\pi }{6}=\frac{1}{2}\] Since,       \[\sin (\alpha +\beta )=1\] \[\Rightarrow \]   \[\sin \,(\alpha +\beta )=\sin \frac{\pi }{2}\]        \[\Rightarrow \]  \[\alpha +\beta =\frac{\pi }{2}\]                               ... (i) and            \[\sin (\alpha -\beta )=\frac{1}{2}\] \[\Rightarrow \]   \[\alpha -\beta =\frac{\pi }{6}\]                 .... (ii) On solving Eqs. (i) and (ii), we get                    \[\alpha =\frac{\pi }{3},\beta =\frac{\pi }{6}\]                     ... (iii) \[\therefore \]     \[\tan \,(\alpha +2\beta )\tan \,(2\alpha +\beta )\]                    \[=\tan \left( \frac{2\pi }{3} \right)\tan \left( \frac{5\pi }{6} \right)\]                                    [putting values from Eq. (iii)]                    \[=\tan \left( \pi -\frac{\pi }{3} \right)\tan \left( \pi -\frac{\pi }{6} \right)\]                    \[=\left( -\cot \frac{\pi }{6} \right)\left( -\cot \frac{\pi }{3} \right)\]                    \[=\sqrt{3}\times \frac{1}{\sqrt{3}}=1\]

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