Prove the following identities :
(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2
Taking LHS = (sin θ + cos θ)2 + (sin θ – cos θ)2
Using the identity,(a + b)2 = (a2 + b2 + 2ab) and (a – b)2 = (a2 + b2 – 2ab)
= sin2 θ + cos2 θ + 2sin θ cos θ + sin2 θ + cos2 θ – 2sin θ cos θ
= 1 +1 [∵ cos2 θ + sin2 θ = 1]
= 2
=RHS
Hence Proved
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