Question

prove: (1-sinA+cosA)^2=2(1+cosA)(1-sinA)​

Answer 1

Answer

Considering LHS: 

(1−sinA+cosA)2=1+sin2A+cos2A−2sinA+2cosA−2sinAcosA

 =2−2sinA+2cosA−2sinAcosA

Considering RHS:

2(1−sinA)(1+cosA)=2(1+cosA−sinA−sinAcosA)

=2+2cosA−2sinA−2sinAcosA

2−2sinA+2cosA−2sinAcosA=2+2cosA−2sinA−2sinAcosA

∴LHS=RHS

Answer 2

$( {1 - sina + cosa})^{2} \\ \\ = ( {1 - sina})^{2} + {cos}^{2} a + 2(1 - sina)cosa \\ \\ = 1 + {sin}^{2} a - 2sina + {cos}^{2} a + 2(1 - sina)cosa \\ \\ = 2 - 2sina + 2(1 - sina)cosa \\ \\ = 2(1 - sina) + 2(1 - sina)cosa \\ \\ = 2(1 - sina)(1 + cosa)$