Question

sin A and cosA are the zeros of polynomial ax2 +bx+c then prove that a2+2ac=b2

Answer 1

if the roots are sinA and cos A,
sinA+cosA=-b/a
sinA.cosA=c/a

(sinA+cosA)² = (-b/a)²
⇒sin²A + cos²A + 2.sinA.cosA = b²/a²
⇒1 + 2(c/a) = b²/a²
⇒ a²×(1 + 2(c/a)) = a²×(b²/a²)    (multiplying a² on both sides)
⇒ a² + 2ac = b²     (proved)