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COMPLEX NUMBERS AND QUADRATIC EQUATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS1 Mark
MCQ
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{c}$ have roots $\alpha$ and $\beta$.Then, what will be the value of $\sin\alpha + \sin\beta$
  • ✓
    $\frac{2\text{bc}}{(\text{a}^2+\text{b}^2)}$
  • B
    $0$
  • C
    $1$
  • D
    $\frac{(\text{c}^2+\text{a}^2)}{(\text{a}^2+\text{b}^2)}$
✓

Answer

Correct option: A.
$\frac{2\text{bc}}{(\text{a}^2+\text{b}^2)}$
Given, $\text{a}\cos\theta+\text{b}\sin\theta=\text{c}$
So, this implies $\text{a}\cos\theta=\text{c}-\text{b}\sin\theta$
Now squaring both the sides we get,
$(\text{a}\cos\theta)^2=(\text{c}-\text{b}\sin\theta)^2$
$\text{a}^2\cos^2\theta=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
$\text{a}^2(1-\sin^2\theta)=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
$\text{a}^2-\text{a}^2\sin^2\theta=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
Now rearranging the elements,
$\text{a}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta+(\text{c}^2-\text{a}^2)=\theta$
So, as sum of the roots are in the form $\frac{-\text{b}}{\text{a}}$ if there is a quadratic equation $\text{a}\text{x}^2=\text{bx}+\text{c}=0$
Now, we can conclude that
$\sin\alpha+\sin\beta=\frac{2\text{bc}}{(\text{a}^2+\text{b}^2)}$

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