The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be
- A) $x^2-2x+2=0$
- B) $x^2-5x+5=0$
- C) $x^2-7x+7=0$
- D) $x^2-9x+9=0$
Method $1$:$$\sec^2\theta+\csc^2\theta=\frac1{\cos^2\theta}+\frac1{\sin^2\theta}\\=\frac1{\sin^2\theta\cos^2\theta}\\=\frac4{\sin^22\theta}\ge4$$
Also, $\sec^2\theta\csc^2\theta=\dfrac1{\sin^2\theta\cos^2\theta}$
So, options $B),C),D)$ are correct.
Method $2$: Let the quadratic equation be $x^2-px+q=0$
So, $\sec^2\theta+\csc^2\theta=p, \sec^2\theta\csc^2\theta=q\implies \csc^2\theta=\dfrac{q}{\sec^2\theta}$
Putting that in the sum of roots, we get $$\sec^2\theta+\frac{q}{\sec^2\theta}=p\\\implies\sec^4\theta-p\sec^2\theta+q=0\\\implies\sec^2\theta=\frac{p\pm\sqrt{p^2-4q}}2\ge1\\\implies p\pm\sqrt{p^2-4q}\ge2\\\implies\pm\sqrt{p^2-4q}\ge2-p\\\implies p^2-4q\ge4+p^2-4p\\\implies p-q\ge1$$
What's wrong in this method?
Best Answer
See my comment under the question and go on from there, as follows: $$ \frac {-p\pm\sqrt{p^2+4p}} 2 = \begin{cases} -1\pm\sqrt3 & \text{if } p=2, \\ \quad\cdots & \text{if } p=5, \\ \quad\cdots & \text{if } p=7, \\ \quad\cdots & \text{if } p=9. \end{cases} $$ When $p=2,$ one of the solutions is negative and so cannot be $\sec^2\theta$ or $\csc^2\theta$ if $\theta$ is real. The other one is negative and so cannot be $\sec^2\theta$ or $\csc^2\theta$ if $\theta$ is real.
Go on from there is a similar way.