Sin^4+ sin^2Acos^2A=sin^2A
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1
Firstly take sin^2A common of LHS
sin^2 A( sin^2A +Cos^2A)
and we know sin^2A+cos^2A=1
Then LHS is sin^2A which is equal to RHS
sin^2 A( sin^2A +Cos^2A)
and we know sin^2A+cos^2A=1
Then LHS is sin^2A which is equal to RHS
Answered by
0
LHS :
sin⁴A + sin²A * cos²A
Take sin²A common,
=> sin²A(sin²A + cos²A)
∴ sin²A + cos²A = 1
=> sin²A(1)
=> sin²A
RHS
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