Q.If sinθ + cosθ = √2 (where 0° < θ < 90°), then the value of θ is (a) 30° (b) 45° (c) 60° (d) 90°
Answer: B
sinθ + cosθ = √2 --- (i)
Or, \((sinθ + cosθ)^2 = 2\)
Or, \(sin^2θ + cos^2θ + 2sinθcosθ = 2\)
Or, \(1 + 2sinθcosθ = 2\)
Or, \(2sinθcosθ = 1\)
Or, \(1 - 2sinθcosθ = 0\)
Or, \(sin^2θ + cos^2θ - 2sinθcosθ = 0\)
Or, \((sinθ - cosθ)^2 = 0\)
Or, \(sinθ - cosθ = 0\) --- (ii)
Adding (i) and (ii), we get \(2sinθ = √2\)
Or, \(sinθ = \cfrac{1}{√2} = sin45°\)
Thus, \(θ = 45°\).
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