Question

If sinθ and Cosθ are roots of the equation ax² + bx + c = 0, Prove that a² - b² + 2ac = 0

Answer 1

$\boxed{\bold{\large{Answer:}}}$

$\text{We\:know\:that,}$

$Sum\:of\:the\:Roots = \frac{ - B}{A} \\ \\ \implies \: \sin \theta \: + \cos \theta = \frac{ - b}{a}$

$\text{Product\:of\:the\:roots}$

$= \frac{C}{A} \\ \\ \implies \: \sin\theta. \cos \theta = \frac{c}{a}$

$\text{Now\:we\:have\:Sin^2\: \theta \:+\:Cos^2\: \theta \:=\:1}$

$\implies ( \frac{ - b}{a} {)}^{2} - 2. \frac{c}{a} = 1 \\ \\ \implies \frac{ {b}^{2} }{ {a}^{2} } - \frac{2c}{a} = 1 \: \: \: or \: \: \: {b}^{2} - 2ac = {a}^{2} \\ \\ \implies {a}^{2} - {b}^{2} + 2ac = 0$