A) \[2n\pi +B\]
B) \[2n\pi -B\]
C) \[n\pi +B\]
D) \[n\pi +B{{(-1)}^{n}}B\]
Correct Answer: A
Solution :
Since sin A = sin B \[\Rightarrow \] \[~A=B\]and\[A=\pi -B\] ...(i) and \[cos\,A=cos\,B\] \[\Rightarrow \] \[A=B\]and\[A=2\pi -B\] ...(ii) From Eqs. (i) and (ii) \[\Rightarrow \] \[A=2n\pi +B\]You need to login to perform this action.
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