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2014-08-14T22:37:44+05:30

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CosA +sinA = √2cosA
squaring
⇒(cosA + sin A)² = (√2cosA)²
⇒cos²A + sin²A + 2sinAcosA = 2cos²A
⇒1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A
⇒2 - 2cos²A = cos²A + sin²A - 2sinAcosA 
⇒2(1 - cos²A)= (cosA - sinA)²
⇒ cosA - sinA = √[2sin²A]
cosA-sinA = √2sinA
hence proved

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2014-08-15T08:25:29+05:30
cos A +sin A = √2cos A
squaring on both sides we get
(cos A + sin A)² = (√2cos A)²
it is in the form of (a+b)^2
formula for (a+b)^2=a^2+ab+b^2
by comparing here we have a=cos A and b=sin A
cos²A + sin²A + 2sinAcosA = 2cos²A
we know that
sin^2A+cos^2 A=1
from this we can write
sin^2 A=1-cos^2 A
and cos^2 A=1-sin^2 A
1 - sin²A + 1 - cos²A + 2 sinAcosA = 2 cos²A
adding like terms and sin^2 A+cos^2 A -2 sinA cosA -2 cos^2 A on both sides
2 - 2 cos²A = cos²A + sin²A - 2 sinAcosA 
taking 2 as common in left side of the equation
2(1 - cos²A)= (cos A - sin A)²
 cos A - sin A = √[2 sin²A]
cos A-sin A = √2 sin A
hence it is proved


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