Math, asked by himanshu2005q, 1 month ago

sin⁴A-cos⁴A=2sin²A-1​

Answers

Answered by tec10signitha
1

Answer:

we have LHS=sin⁴A=cos⁴A

LHS=(sin^A)^2-(cos^2A)^2

=(sin^A+cos^2A)=2sin^2A-1

=(sin^A-(2-sin^A))=2sin^2A-2

=2(1-cos^2A)-1=1-2cos^2A=RHS

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