Question

if sin theta and Cos theta are the roots of equation X square + bx + c equal to zero prove that a square minus b square + 2 is equal to zero ​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

Cosθ and sinθ are roots of equation x² + bx + c = 0

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

$\large \star {\boxed{\sf{Sum \: of \: zeros \: = \: \dfrac{-b}{a}}}} \\ \\ \implies {\sf{ \cos \theta \: + \: \sin \theta \: = \: \dfrac{-b}{a}}} \\ \\ \small{\underline{\pink{\sf{ \: \: \: \: \: \: \: \: \: \: Square \: Both \: Sides \: \: \: \:!\: \: \: \: \: \: }}}}$

$\implies {\sf{\big( \cos \theta \: + \: \sin \theta \big)^2 \: = \: \big( \dfrac{-b}{a} \big)^2 }} \\ \\ \implies {\sf{ \cos^2 \theta \: + \: \sin^2 \theta \: + \: 2 \cos \theta \sin \theta \: = \: \dfrac{b^2}{a^2}}} \\ \\ \implies {\sf{ 1 \: + \: 2 \cos \theta \sin \theta \: = \: \dfrac{b^2}{a^2}---(1)}}$

And for product we have formula :

$\large \star {\boxed{\sf{Product \: = \: \dfrac{c}{a}}}} \\ \\ \implies {\sf{ \sin \theta \cos \theta \: = \: \dfrac{c}{a}}} \\ \\ \implies {\sf{1 \: + \: 2 \dfrac{c}{a} \: = \: \dfrac{b^2}{a^2}}} \\ \\ \small {\underline{\pink{\sf{\: \: \: \: \: \: \: \: \: Multiply \: by \: a^2 \: \: \: \: \: \: \: \: }}}} \\ \\ \implies {\sf{a^2 \: + \: 2ac \: = \: b^2}} \\ \\ \implies {\sf{a^2 \: - \: b^2 \: + \: 2ac \: = \: 0}}$

$\large {\mathbb{HENCE \: \: PROVED}}$

Answer 2

Correct Question

If sin∅ and cos∅ are the roots of equation ax² + bx + c = 0, prove that a² - b² + 2ac = 0

Solution

We know that sum of zeroes of a quadratic equation

= -b/a

→ sin∅ + cos∅ = -b/a

Squaring both sides, we get

(sin∅ + cos∅)² = (-b/a)²

→ sin²∅ + cos²∅ + 2sin∅cos∅ = b²/a²

(using, (a + b)² = a² + b² + 2ab)

We know that sin²∅ + cos²∅ = 1

Hence the equation becomes

→ 1 + 2sin∅cos∅ = b²/a²

We know that product of zeroes in a quadratic equation = c/a

→ sin∅cos∅ = c/a

Hence,

1 + 2c/a = b²/a²

Multiplying both sides by a², we get

→ a² + 2ac = b²

→ a² - b² + 2ac = 0

Hence Proved.