• # 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

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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