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Question

Prove that :
sinA+cosAsinAcosA+sinAcosAsinA+cosA=2sin2Acos2A=22sin2A1=212cos2A

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Solution

L.H.S. = sinA+cosAsinAcosA+sinAcosAsinA+cosA

L.H.S. = (sinA+cosA)2+(sinAcosA)2(sinAcosA)(sinA+cosA)

L.H.S. = (sin2A+cos2A+2sinAcosA)+(sin2A+cos2A2sinAcosA)sin2Acos2A

L.H.S. = (1+2sinAcosA)+(12sinAcosA)sin2Acos2A

L.H.S. = 2sin2Acos2A=2sin2A(1sin2A)

L.H.S. = 22sin2A1=22(1cos2A)1=212cos2A=R.H.S.

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