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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