Question

sin⁴A+cos⁴A=1-2sin²A×cos²A

Answer 1

sin⁴A + cos⁴A = 1 - 2sin²Acos²A
LHS
(sin²A)² + (cos²A)² = ( sin²A + cos²)²
= (sin²A + cos²A) - 2(sin²A×cos²A
= 1 - 2sin²Acos²A = RHS [Proved]

Answer 2

sin⁴A+cos⁴A=1-2sin²A×cos²A

On solving LHS

On solving LHS(Sin²A)²+ (Cos²A)²

On solving LHS(Sin²A)²+ (Cos²A)² (Sin²A+Cos²A)² by using [ (a+b)²=a²+b²+2ab]

On solving LHS(Sin²A)²+ (Cos²A)² (Sin²A+Cos²A)² by using [ (a+b)²=a²+b²+2ab](Sin²A)²+(Cos²A)²+2Sin²ACos²A

On solving LHS(Sin²A)²+ (Cos²A)² (Sin²A+Cos²A)² by using [ (a+b)²=a²+b²+2ab](Sin²A)²+(Cos²A)²+2Sin²ACos²A(Sin²A+Cos²A)²+2Sin²ACos²A

On solving LHS(Sin²A)²+ (Cos²A)² (Sin²A+Cos²A)² by using [ (a+b)²=a²+b²+2ab](Sin²A)²+(Cos²A)²+2Sin²ACos²A(Sin²A+Cos²A)²+2Sin²ACos²A(1)²+2Sin²ACos²A by using [ ( Sin²A+Cos²A=1) ]

On solving LHS(Sin²A)²+ (Cos²A)² (Sin²A+Cos²A)² by using [ (a+b)²=a²+b²+2ab](Sin²A)²+(Cos²A)²+2Sin²ACos²A(Sin²A+Cos²A)²+2Sin²ACos²A(1)²+2Sin²ACos²A by using [ ( Sin²A+Cos²A=1) ]

On solving LHS(Sin²A)²+ (Cos²A)² (Sin²A+Cos²A)² by using [ (a+b)²=a²+b²+2ab](Sin²A)²+(Cos²A)²+2Sin²ACos²A(Sin²A+Cos²A)²+2Sin²ACos²A(1)²+2Sin²ACos²A by using [ ( Sin²A+Cos²A=1) ] 1-2Sin²A×Cos²A

THEREFORE :- LHS=RHS

THEREFORE :- LHS=RHSHENCE VERIFIED