Proved it
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Given,
cosθ + sinθ= √2cosθ
squaring on both the sides, we get,
cos^2 θ+ sin^2 θ + 2sinθ cosθ = 2cos^2 θ
cos^2 θ − sin^2 θ = 2cos θ sin θ
(cosθ+sinθ) (cosθ−sinθ) = 2cos θ sinθ
√2cos θ (cosθ−sinθ) = 2cos θ sin θ
[ Given cosθ+sinθ= √2cosθ]
∴ cos θ − sin θ = √2 sin θ
Hence proved
Bipasha2007:
thanks
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