prove that sin⁴A-cos⁴A=2sin²A-1
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Step-by-step explanation:
LHS. RHS
Sin⁴A-Cos⁴A. 2Sin²A-1
Bringing -1 from RHS to LHS, we get LHS as
Solving LHS
=Sin⁴A-Cos⁴A+1
= (sin²A)²-(cos²A)²+1
Using Identity a²-b²= (a-b)(a+b)
(sin²A+cos²A)(sin²A-cos²A)+1
We know that
sin²A+cos²A = 1
So we get the result as
sin²A-cos²A+1
We know that
-cos²A+1 = sin²A OR 1 - cos²A = sin²A
Placing Sin²A we get
Sin²A+ sin²A
= 2sin²A
RHS
2Sin²A
So LHS = RHS
Hence Proved.
Identities used
a²-b²= (a-b)(a+b)
sin²A+cos²A = 1
-cos²A+1 = sin²A
Or 1 - cos²A = sin²A
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