sin^4x+cos^4x=1-2sin^2xcos^2x
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Given Sin⁴x+Cos⁴x=1-2Sin²xCos²x
⇒(Sin²x)²+(Cos²x)²=1-2Sin²xCos²x
⇒(Sin²x)²+(Cos²x)²+2Sin²xCos²x=1
⇒(Sin²x+Cos²x)²=1
⇒(1)²=1
⇒1=1
Therefore,LHS=RHS.
⇒(Sin²x)²+(Cos²x)²=1-2Sin²xCos²x
⇒(Sin²x)²+(Cos²x)²+2Sin²xCos²x=1
⇒(Sin²x+Cos²x)²=1
⇒(1)²=1
⇒1=1
Therefore,LHS=RHS.
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