A) \[0{}^\circ \]
B) \[30{}^\circ \]
C) \[45{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: A
Solution :
Given, \[\sin \,2A=2\,\sin A\] when, \[A=0{}^\circ \], then \[\sin \,\left( 2\times 0{}^\circ \right)=2\sin \,0{}^\circ \] \[\Rightarrow \,\,\sin \left( 0{}^\circ \right)=2\times 0\] \[\Rightarrow 0=0\] So, for \[A=0{}^\circ \], given statement is true.
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