If sinθ and cosθ are roots of the equation ax2+bx+c=0, prove that . . .

Question : If sinθ and cosθ are roots of the equation ax²+bx+c=0, prove that a²-b²+2ac=0. 

Doubt by Charu 

Solution : 
ax²+bx+c=0
α=sinθ (Given)

β=cosθ (Given)

Sum of Zeroes

α+β = -b/a
sinθ+cosθ = -b/a — (1)

Product of Zeroes
αβ = c/a
sinθ.cosθ = c/a — (2)

We know, 

α²+β² = (α+β)²-2αβ

sin²θ + cos²θ = (sinθ+cosθ)²-2sinθ.cosθ
1 = (-b/a)²-2(c/a)
[sin²θ+cos²θ=1] [From eq (1) and (2)]

1=b²/a² - 2c/a
1=(b²-2ac)/a²
a²=b²-2ac
a²-b²+2ac=0

Hence Proved.