If sin θ and cos θ are the roots of the equations ax2 – bx + c = 0

1.	If sin θ and cos θ are the roots of the equations ax2 – bx + c = 0, then which of the following is correct?(a) a2 + b2 + 2ac = 0 (b) a2 – b2 + 2ac = 0 (c) a2 + b2 + 2ab = 0 (d) a2 – b2 – 2ac = 0.
Answer» (b) a² – b² + 2ac = 0. As sin θ and cos θ are the roots of the equation ax² – bx + c = 0. ∴ sin θ + cos θ = \(\frac{b}{a}\) and sin θ cos θ = \(\frac{c}{a}\) ⇒ (sin θ + cos θ)² = \(\frac{b^2}{a^2}\) ⇒ sin²θ + cos²θ + 2sin θ + cos θ = \(\frac{b^2}{a^2}\) ⇒ 1 + \(\frac{2c}{a}=\) \(\frac{b^2}{a^2}\) ⇒ \(\frac{2c}{a}=\)\(\frac{b^2}{a^2}\) - 1 = \(\frac{b^2-a^2}{a^2}\) ⇒ 2ac = b² – a² ⇒ a² – b² + 2ac = 0.

If sin θ and cos θ are the roots of the equations ax2 – bx + c = 0, then which of the following is correct?(a) a2 + b2 + 2ac = 0 (b) a2 – b2 + 2ac = 0 (c) a2 + b2 + 2ab = 0 (d) a2 – b2 – 2ac = 0.

Answer»

(b) a² – b² + 2ac = 0.

As sin θ and cos θ are the roots of the equation ax² – bx + c = 0.

∴ sin θ + cos θ = \(\frac{b}{a}\) and sin θ cos θ = \(\frac{c}{a}\)

⇒ (sin θ + cos θ)² = \(\frac{b^2}{a^2}\)

⇒ sin²θ + cos²θ + 2sin θ + cos θ = \(\frac{b^2}{a^2}\)

⇒ 1 + \(\frac{2c}{a}=\) \(\frac{b^2}{a^2}\) ⇒ \(\frac{2c}{a}=\)\(\frac{b^2}{a^2}\) - 1 = \(\frac{b^2-a^2}{a^2}\)

⇒ 2ac = b² – a² ⇒ a² – b² + 2ac = 0.