Question

show that (1-sina+cosa)^2 =2(1+cosa)(1-sina)

Answer 1

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

Hope it helps.

Answer 2

Answer:

(1-sinA +cosA)^2=2(1+cosA)(1-sinA)

Step-by-step explanation:

LHS=(1-sinA+cosA)^2

=[(1-sinA)+cosA]^2

=(1-sinA)^2+cos^A+2(1-sinA)cosA

=1+sin^2A-2sinA+cos^2A+2(1-sinA)cosA

=1+(sin2A+cos^2A)-2sinA+2(1-sinA)cosA

=1+1-2sinA+2(1-sinA)cosA

=2(1-sinA)+2(1-sinA)cosA

=2(1-sinA)(1+cosA)

=RHS