Question

If $\sin \theta $ and $\cos \theta $ are the roots of the equation $a{x^2} - bx + c = 0$ , then $a,b{\text{ and c}}$ satisfy the relation
$
  {\text{A}}{\text{. }}{a^2} + {b^2} + 2ac = 0 \\
  {\text{B}}{\text{. }}{a^2} - {b^2} + 2ac = 0 \\
  {\text{C}}{\text{. }}{a^2} + {b^2} + 2ab = 0 \\
  {\text{D}}{\text{. }}{a^2} - {b^2} - 2ac = 0 \\
 $

Answer 1

Hint: From this quadratic equation, first we need to find the sum and product of roots of the equation and then by applying some trigonometric identities we solve the question further. By using the simple algebra, we will get our answer.

Complete step-by-step answer:
Given that $\sin \theta $ and $\cos \theta $ are the roots of the equation $a{x^2} -bx + c = 0$
We know sum of roots = $\dfrac{-b}{a}$ and product of roots = $\dfrac{c}{a}$ for quadratic equation $a{x^2} +bx + c = 0$
So when we find sum and product of roots comparing
$
  \sin \theta + \cos \theta = \dfrac{b}{a}.....(1) \\
  \sin \theta \times \cos \theta = \dfrac{c}{a} \\
 $
Then by squaring equation (1) on both sides
$
  {\left( {\sin \theta + \cos \theta } \right)^2} = \dfrac{{{b^2}}}{{{a^2}}} \\
  {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = \dfrac{{{b^2}}}{{{a^2}}} \\
 $
Now substituting the values from product equation and applying the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$1 + \dfrac{{2c}}{a} = \dfrac{{{b^2}}}{{{a^2}}}$
Now solving both sides
$\dfrac{{a + 2c}}{a} = \dfrac{{{b^2}}}{{{a^2}}}$
Now, by cross multiplying
$
  {a^2} + 2ac = {b^2} \\
  {a^2} - {b^2} + 2ac = 0 \\
 $
Hence option ${\text{B}}{\text{. }}{a^2} - {b^2} + 2ac = 0$ is the right answer.

Note: You need to remember the trigonometric identities for sine and cosine to solve this question. Carefully, find the sum and product and obtain the correct answer using the method mentioned above. Apply the formulas and cross multiply with caution as these steps are most error prone. And in the question of quadratic equations this is the only method to achieve our answer.