Q3 of 80 Page 311

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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