A) \[\sin \alpha \]
B) \[\cos \alpha \]
C) \[\sin \beta \]
D) \[\cos 2\beta \]
Correct Answer: A
Solution :
Given, cot \[(\alpha +\beta )=0\Rightarrow \cos (\alpha +\beta )=0\] Þ \[\alpha +\beta =(2n+1)\frac{\pi }{2},n\in I\] \\[\sin (\alpha +2\beta )=\sin (2\alpha +2\beta -\alpha )\]=\[\sin \text{ }[(2n+1)\text{ }\pi -\alpha ]\] \[=\sin (\,2n\pi +\pi -\alpha \,)\] = \[\sin (\,\pi -\alpha \,)\,=\sin \alpha \].
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