Question

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

• A. $p^2-q^2+2pr=0$
• B. $(p+r{)}^2=q^2-r^2$
• C. $p^2+q^2-2pr=0$
• D. $(p-r{)}^2=q^2+r^2$

Answer 1

The correct option is A $p^2-q^2+2pr=0$

${px}^2+qx+r=0$

roots are $\sin \theta$ and $\cos \theta$

$\sin \theta +\cos \theta =\frac{-q}{p}$ $\sin \theta .\cos \theta =r/p$

${(\sin \theta +\cos \theta )}^2=\frac{q^2}{p^2}$

$1+2\left(\frac{r}{p}\right)=\frac{q^2}{p^2}$

$\frac{p+2r}{p}=\frac{q^2}{p^2}$

$p(p+2r)=q^2$

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