If
then let us write by calculating, the value of (sin θ + cos θ).
⇒ ![]()
⇒ 2sinθcosθ = 1
⇒ we know that, sin2θ + cos2θ = 1
Can be written as
⇒ (sinθ + cosθ)2 – 2sinθcosθ = 1
⇒ (sinθ + cosθ)2-1 = 1
⇒ (sinθ + cosθ)2 = 2
⇒ (sinθ + cosθ) = √2
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