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if cosA +sinA=root2cosA show that cosA -sinA=root2 sina

cos A + sin A = √2 cos A  .............(1)

Squaring both sides , we get

⇒ ( cos A + sin A )2 = 2 cos2A

⇒ cos2 A + sin 2A + 2 sin A cos A =  2 cos2A 

⇒ cos2 A - sin 2A = 2 sin A cos A

⇒ ( cos A + sin A) ( cos A - sin A) = 2 sin A cos A  

⇒√2 cos A  (cos A - sin A) = 2 sin A cos A  [Using (1) ]

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i think the question shuld b lyk this:

IF cosA-sinA=root2sinA

PROVE THAT: cosA+sinA=root2 cosA

cosA - sinA = (√2)sinA

=> (cosA - sinA)2 = [(√2)sinA]2 [Squaring both sides]

=> cos2A - 2cosAsinA + sin2A = (√2)2*sin2A

=> (cos2A + sin2A) - 2cosAsinA = 2sin2A

=> 1 - 2cosAsinA = 2(1 - cos2A)

=> 1 - 2cosAsinA = 2 - 2cos2A

=> (1 - 2cosAsinA) - 2 = -2cos2A

=> -1 - 2cosAsinA = -2cos2A

=> -1 * ( -1 - 2cosAsinA ) = -1 * (-2cos2A) [Multiplying both sides by (-1)]

=> 1 + 2cosAsinA = 2cos2A

=> (cos2A + sin2A) + 2cosAsinA = 2cos2A

=> cos2A + 2cosAsinA + sin2A = 2cos2A

=> (cosA + sinA)2 = 2cos2A

=> cosA + sinA = √(2cos2A)

=> cosA + sinA = (√2)cosA

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