Q. 1 B4.1( 14 Votes )

# Evaluate the following :

B. (sinθ + cosθ)^{2} + (sinθ - cosθ)^{2}

Answer :

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) [∵, a^{2} + b^{2} = (a + b)^{2} – 2ab]

⇒ (sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} = (sin θ + cos θ + sin θ – cos θ)^{2} – 2(sin^{2} θ – cos^{2} θ)

[∵, (a + b)(a – b) = a^{2} – b^{2}]

⇒ (sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} = (2 sin θ)^{2} – 2 sin^{2} θ + 2 cos^{2} θ

⇒ (sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} = 4 sin^{2} θ – 2 sin^{2} θ + 2 cos^{2} θ

⇒ (sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} = 2 sin^{2} θ + 2 cos^{2}θ

⇒ (sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} = 2 (sin^{2} θ + cos^{2} θ)

⇒ (sin θ + cos θ)^{2} + (sin θ – cos θ)^{2} = 2 [∵, sin^{2} θ + cos^{2} θ = 1]

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

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