If sin θ and cos θ are the roots of the equation ax² − bx + c = 0, then a, b and c satisfy the relation(a) a² + b² + 2ac = 0(b) a² − b² + 2ac = 0(c) a² + c² + 2ab = 0(d) a² − b² − 2ac = 0

Question

\[ \text{If } \sin\theta \text{ and } \cos\theta \]

\[ \text{are the roots of the equation } ax^2-bx+c=0, \]

\[ \text{then } a,b \text{ and } c \text{ satisfy the relation} \]

(a) \(a^2+b^2+2ac=0\)
(b) \(a^2-b^2+2ac=0\)
(c) \(a^2+c^2+2ab=0\)
(d) \(a^2-b^2-2ac=0\)

Since roots are \(\sin\theta\) and \(\cos\theta\),

\[ \sin\theta+\cos\theta=\frac{b}{a} \]

\[ \sin\theta\cos\theta=\frac{c}{a} \]

Using identity

\[ (\sin\theta+\cos\theta)^2 = \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta \]

\[ = 1+2\sin\theta\cos\theta \]

Substituting values,

\[ \left(\frac{b}{a}\right)^2 = 1+2\left(\frac{c}{a}\right) \]

\[ \frac{b^2}{a^2} = \frac{a^2+2ac}{a^2} \]

\[ b^2=a^2+2ac \]

\[ a^2-b^2+2ac=0 \]

\[ \boxed{a^2-b^2+2ac=0} \]

Correct Option: (b)

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