Zane Decker

2022-04-08

The quadratic equation whose roots are $\mathrm{sec}2\theta$ and $\mathrm{csc}\theta$ can be
a) ${x}^{2}-2x+2=0$
b) ${x}^{2}-5x+5=0$
c) ${x}^{2}-7x+7=0$
d) ${x}^{2}-9x+9=0$

Kendall Clark

Step 1
As others have pointed out, you’ve assume that if $u\ge v$ then ${u}^{2}\ge {v}^{2}$
This is true if u,v are both non-negative, but if both are non-positive, then the inequality reverses, ${u}^{2}\le {v}^{2}$, and there is nothing we can say if u is positive and v negative.
Another way to write the equation is as
$\left(\left({\mathrm{cos}}^{2}\theta \right)x-1\right)\left(\left({\mathrm{sin}}^{2}\theta \right)x-1\right)=\left({\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta \right){x}^{2}-x+1$
So the monic polynomial with these two roots has
$p=q=\frac{1}{{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=\frac{4}{{\mathrm{sin}}^{22}\theta }$
This is really the same as your first approach, but switches to sin and cos immediately.

Kendall Wilkinson

Step 1
Recall that
$\frac{1}{{\mathrm{sec}}^{2}\theta }+\frac{1}{{\mathrm{csc}}^{2}\theta }=1.$ If
${x}^{2}+px+q=0$
then
$x={x}_{1}=\frac{-p+\sqrt{{p}^{2}-4q}}{2}$
or
$x={x}_{2}=\frac{-p-\sqrt{{p}^{2}-4q}}{2}.$
So you want $\frac{1}{{x}_{1}}+\frac{1}{{x}_{2}}=1$ You have
$\frac{1}{{x}_{1}}+\frac{1}{{x}_{2}}=\frac{2}{-p+\sqrt{{p}^{2}-4q}}+\frac{2}{-p-\sqrt{{p}^{2}-4q}}=\frac{-p}{q}$
So,
$1=\frac{-p}{q}$
$p=-q$

When $p=2$, one of the solutions is negative and so cannot be $\mathrm{sec}2\theta$ or $\mathrm{csc}2\theta$ if $\theta$ is real. The other one is negative and so cannot be $\mathrm{sec}2\theta$ or $\mathrm{csc}2\theta$ if $\theta$ is real.
Go on from there is a similar way.

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