Question

If $\sin \alpha$and $\cos \alpha$ are the roots of the equation $p{x}^2+qx+r=0$ then

• A. $p^2+q^2-2pr=0$
• B. $p^2-q^2+2pr=0$
• C. $p^2-q^2-2pr=0$
• D. $p^2+q^2+2pr=0$

Answer 1

The correct option is B

$p^2-q^2+2pr=0$

Explanation for the correct option.

Step 1: Simplify the equation

Given that, $p{x}^2+qx+r=0$ , $\sin \alpha$ and $\cos \alpha$ are the roots.

We know that, Sum of the roots $=\sin \alpha +\cos \alpha =-\frac{[\mathrm{coefficient}\mathrm{of}\mathrm{x}]}{[\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2]}$

$\Rightarrow \sin \alpha +\cos \alpha =-\frac{q}{p}.......\left(1\right)$
Product of the roots $=\sin \alpha \cos \alpha$

$=\frac{r}{p}....\left(2\right)$
$\begin{array}{rcl}{\sin }^2\alpha +{\cos }^{2}a& =& 1\\ & \Rightarrow & {\sin }^2\alpha +{\cos }^2\alpha +2\cos \alpha \sin \alpha -2\cos \alpha \sin \alpha =1\\ \Rightarrow (\sin \alpha +\cos \alpha {)}^2-2\cos \alpha \sin \alpha & =& 1\end{array}$

Step 2: Determine the relation

Substituting values from (1) & (2) we get:
$\begin{array}{rcl}{\left(-\frac{q}{p}\right)}^2-2\left(\frac{r}{p}\right)& =& 1\\ \frac{q^2}{p^2}-\frac{2r}{p}& =& 1\\ \frac{\left[q^2-2pr\right]}{p^2}& =& 1\\ \left[q^2-2pr\right]& =& p^2\\ p^2-q^2+2pr& =& 0\end{array}$
Hence, option B is correct.