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To prove : sin⁴A + sin²Acos²A = sin²A
Solving left hand side,
= > sin⁴A + sin²A cos²A
= > ( sin²A )² + sin²A cos²A
= > ( sin²A x sin²A ) + sin²A cos²A
= > sin²A { sin²A + cos²A }
==============
From the properties of trigonometry, we know,
sin²A + cos²A = 1
==============
Therefore,
= > sin²A( 1 )
= > sin²A x 1
= > sin²A
= > RHS
Thus,
LHS = RHS
sin⁴A + sin²Acos²A = sin²A
Proved.
Solving left hand side,
= > sin⁴A + sin²A cos²A
= > ( sin²A )² + sin²A cos²A
= > ( sin²A x sin²A ) + sin²A cos²A
= > sin²A { sin²A + cos²A }
==============
From the properties of trigonometry, we know,
sin²A + cos²A = 1
==============
Therefore,
= > sin²A( 1 )
= > sin²A x 1
= > sin²A
= > RHS
Thus,
LHS = RHS
sin⁴A + sin²Acos²A = sin²A
Proved.
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