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SinA+ cos A=√2sinA then,sinA-cosA= |
Answer» Given :To find :
Solution :Sin A + Cos A = √2 sin A
(sin A + cos A)² = (√2 sin A)²
Sin² A + Cos² A + 2SIN A Cos A = 2 sin²A ____(1) Using Identity Sin² A + cos ² A = 1 Sin² A = 1 - cos² A ______(2) Using equation 2 JININ n equation 1 : 1 - cos² A + cos² A + 2 sin A cos A = 2 ( 1 - cos²A) 1 + 2 sin A cos A = 2 - 2 cos² A 2 sin A cos A = 2 - 1 - 2 cos² A 2 sin A cos A = 1 - 2cos² A 2cos² A = 1 - 2 sin A cos A 2Cos² A = sin² A + cos² A - 2 sin A cos A Now , Using identity ( x - y)² = x² + y² - 2xy 2Cos² A = ( sin A - cos A)² Sin A - cos A = √(2 cos² A) Sin A - cos A = ± √2 cos A |
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