InterviewSolution
Saved Bookmarks
| 1. |
sina-2sin3a/2cos3a-cosa=tana |
|
Answer» c LHS=sinA(1-2sin2A) / cos A (2cos2A-1)=\xa0sinA(sin2A + cos2A-2sin2A) / cos A (2cos2A-(sin2A + cos2A))=sinA( cos2A-sin2A) / cosA ( cos2A-sin2A)=sinA / cosA = tanA=RHS |
|