if sin0 and cos0 are the roots of the equation ax2 + bx +c=0 , prove that a2-b2 +2ab=0.
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Answers
Correct Question :
If sin ∅ and cos ∅ are the roots of the equation ax² + bx + c = 0, prove that a² - b² + 2ac = 0.
Answer :
Given :
sin ∅ and cos ∅ are the roots of the equation ax² + bx + c = 0
We know that
Sum of roots of the given = sin ∅ + cos ∅ = - coefficient of x / Coefficient of x² = - b/a → Eq( 1 )
Product roots of the given equation = sin ∅.cos ∅ = Constant / Coefficient of x² = c/a → Eq( 2 )
Taking eq( 1 ) and by squaring on both sides we get,
⇒ sin ∅ + cos ∅ = - b/a
⇒ ( sin ∅ + cos ∅ )² = ( - b/a )²
Using algebraic identity ( p + q )² = p² + q² + 2pq
⇒ sin² ∅ + cos² ∅ + 2sin ∅.cos ∅ = b²/a²
Since sin² ∅ + cos² ∅ = 1
⇒ 1 + 2sin ∅.cos ∅ = b²/a²
From eq( 2 )
⇒ 1 + 2c/a = b²/a²
⇒ ( a + 2c )/a = b²/a²
⇒ a + 2c = b²/a
⇒ a( a + 2c ) = b²
⇒ a² + 2ac = b²
⇒ a² - b² + 2ac = 0
Hence Proved.
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