Question

sinA+cos A=√2cos(45°-A)​

Answer 1

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Here's your answer

Given, sinA + cosA = √2cos(45° - A)

RHS = √2cos(45° - A) = √2[cos45°cosA + sin45°sinA] = √2[cosA/√2 + sinA/√2] = √2/√2[cosA + sinA] = sinA + cosA = LHS

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