If sin α and cos α are the roots of the equation ax2 + bx + c = 0, t

1.	If sin α and cos α are the roots of the equation ax2 + bx + c = 0, then b2 = A. a2 – 2ac B. a2 + 2ac C. a2 – ac D. a2 + ac
Answer» Equation ax² + bx + c = 0 has cos a and cos a as two roots sin α + cos α = \(-\frac{b}{a}\) sin α × cos α = c/a …eq(1) \((sin\,a\,+\,cosa)^2=\frac{b^2}{a^2}\) \(sin^2a\,+cos^2a\,+2\,sina\,cos=\frac{b^2}{a^2}....eq(2)\) But sin2 α + cos2 α = 1 ∴ a² (1 + 2 sinα.cos α) = b² Putting sin α × cos α = c/a, we get, ⇒ b²= a²+ 2ac.

If sin α and cos α are the roots of the equation ax2 + bx + c = 0, then b2 = A. a2 – 2ac B. a2 + 2ac C. a2 – ac D. a2 + ac

Answer»

Equation ax² + bx + c = 0 has cos a and cos a as two roots

sin α + cos α = \(-\frac{b}{a}\)

sin α × cos α = c/a …eq(1)

\((sin\,a\,+\,cosa)^2=\frac{b^2}{a^2}\)

\(sin^2a\,+cos^2a\,+2\,sina\,cos=\frac{b^2}{a^2}....eq(2)\)

But sin2 α + cos2 α = 1

∴ a² (1 + 2 sinα.cos α) = b²

Putting sin α × cos α = c/a, we get,

⇒ b²= a²+ 2ac.