sin⁴A-cos⁴A=2sin²A-1
Answers
Answered by
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
Similar questions