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cos ϴ + sin ϴ = √2 cos ϴ
Squaring both sides
(cos ϴ + sin ϴ)² = ( √2 cos ϴ)²
cos²ϴ + 2 cosϴ sinϴ + sin²ϴ = 2 cos²ϴ
cos²ϴ - 2 cos²ϴ + 2 cos ϴ sin ϴ = - sin²ϴ
- ( cos²ϴ - 2 cos ϴ sin ϴ) = - sin²ϴ
cos²ϴ - 2 cos ϴ sin ϴ) = sin²ϴ
cos²ϴ - sin²ϴ = 2 cos ϴ sin ϴ
Using the identity of a² - b²
( cos ϴ + sin ϴ) (cos ϴ - sin ϴ) = 2 cos ϴ sin ϴ
Given, cosϴ - sinϴ = √2 cos ϴ
2 sinϴ cos ϴ = (√2 cos ϴ) (cos ϴ - sin ϴ)
2 sinϴ cosϴ / √2 cosϴ = cos ϴ - sin ϴ
By rationalising, We get,
√2 sin ϴ = cos ϴ - sin ϴ
Hence,Proved.
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