Question

if sin0 and cos0 are the roots of the equation ax2 + bx +c=0 , prove that a2-b2 +2ab=0.
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Answer 1

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.

Answer 2

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