(1-sinA + CosA)² =2(1-sinA)(1+CosA)
Answers
Answered by
2
hiii dear
ur answer
Consider the LHS: (1-sinA+cosA)2 = [(1-sinA) + cosA]2 = (1-sinA)2 + cos2A + 2(1-sinA)cosA = 1 + sin2A − 2sinA + cos2A + 2(1-sinA)cosA = 1 + (sin2A + cos2A) − 2sinA + 2(1-sinA)cosA = 1 + 1 − 2sinA + 2(1-sinA)cosA [Since, sin2A + cos2A =1] = 2 − 2sinA + 2(1-sinA)cosA = 2(1 − sinA) + 2(1-sinA)cosA = 2(1 − sinA)(1 + cosA) = RHS
ur answer
Consider the LHS: (1-sinA+cosA)2 = [(1-sinA) + cosA]2 = (1-sinA)2 + cos2A + 2(1-sinA)cosA = 1 + sin2A − 2sinA + cos2A + 2(1-sinA)cosA = 1 + (sin2A + cos2A) − 2sinA + 2(1-sinA)cosA = 1 + 1 − 2sinA + 2(1-sinA)cosA [Since, sin2A + cos2A =1] = 2 − 2sinA + 2(1-sinA)cosA = 2(1 − sinA) + 2(1-sinA)cosA = 2(1 − sinA)(1 + cosA) = RHS
ayush84340:
not clear
Similar questions