Question

If $\cos \alpha$ and $\sin \alpha$ are the roots of the quadratic equation $p{x}^2+qx+r=0$, then the value of $p^2-q^2+2pr$ is equal to

Answer 1

Sum of roots
$\cos \alpha +\sin \alpha =-\frac{q}{p}\cdots \left(1\right)$
Product of roots
$\cos \alpha .\sin \alpha =\frac{r}{p}\cdots \left(2\right)$
On squaring both sides of the equation $\left(1\right)$,
$(\cos \alpha +\sin \alpha {)}^2={(-\frac{q}{p})}^2$
$\Rightarrow {\cos }^2\alpha +{\sin }^2\alpha +2\cos \alpha .\sin \alpha =\frac{q^2}{p^2}$
$\Rightarrow 1+2\cos \alpha .\sin \alpha =\frac{q^2}{p^2}$
Using equation $\left(2\right)$,
$\Rightarrow 1=\frac{q^2}{p^2}-\frac{2r}{p}$
$\Rightarrow p^2=q^2-2pr$
$\Rightarrow p^2-q^2+2pr=0$