Q. 1 B4.1( 14 Votes )

# Evaluate the following :B. (sinθ + cosθ)2 + (sinθ - cosθ)2

We have

(sin θ + cos θ)2 + (sin θ – cos θ)2 = ((sin θ + cos θ) + (sin θ – cos θ))2 – 2(sin θ + cos θ)(sin θ – cos θ) [, a2 + b2 = (a + b)2 – 2ab]

(sin θ + cos θ)2 + (sin θ – cos θ)2 = (sin θ + cos θ + sin θ – cos θ)2 – 2(sin2 θ – cos2 θ)

[, (a + b)(a – b) = a2 – b2]

(sin θ + cos θ)2 + (sin θ – cos θ)2 = (2 sin θ)2 – 2 sin2 θ + 2 cos2 θ

(sin θ + cos θ)2 + (sin θ – cos θ)2 = 4 sin2 θ – 2 sin2 θ + 2 cos2 θ

(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 sin2 θ + 2 cos2θ

(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 (sin2 θ + cos2 θ)

(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 [, sin2 θ + cos2 θ = 1]

Thus, (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2.

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