The quadratic equation whose roots are $\sec^2\theta$ and $\csc^2\theta$ can be 
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?
 A: 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.
A: As others have pointed out, you’ve assume that if $u\geq v$ then $u^2\geq v^2.$
This is true if $u,v$ are both non-negative, but if both are non-positive, then the inequality reverses, $u^2\leq v^2,$ and there is nothing we can say if $u$ is positive and $v$ negative.
Another way to write the equation is as $$((\cos^2 \theta) x-1)((\sin^2\theta) x-1)=(\cos^2\theta\sin^2 \theta) x^2-x+1$$
So the monic polynomial with these two roots has $$p=q=\frac1{\sin^2\theta\cos^2\theta}=\frac{4}{\sin^22\theta}$$
This is really the same as your first approach, but switches to $\sin$ and $\cos$ immediately.
A: It seems that what we are asked to notice is that
$$ \sec^2 x \ + \ \csc^2 x \ \ = \ \ \frac{1}{\cos^2 x} \ + \ \frac{1}{\sin^2 x} \ \ = \ \ \frac{\sin^2 x \ + \ \cos^2 x}{\cos^2 x·\sin^2 x} \ \ = \ \  \sec^2 x  ·  \csc^2 x \ \ , $$
a perhaps not overly-familiar trigonometry identity.  This is why all of the polynomials in the question choices have the form $ \ x^2  \ - \  m·x  \  + \ m \ \ . $ Since we are apparently to assume that $ \ \theta \ $ is real, we must have of course $ \ \sec^2 \theta \ge 1 \ , \ \csc^2 \theta \ge 1 \ \ . $  But for the equation $ \ x^2    -    m x     +   m \ = \ 0 \ \ , $ this is not possible unless the quadratic polynomial is at least a "binomial-square" $ \ (x - p)^2 \ = \ x^2 - 2px + p^2 \ \ , $ which requires $ \ p^2 \ = \ 2p \ \Rightarrow \ p \ = \ 2 \ \Rightarrow \ m \ = \ 4 \ \ . $  Since choice $ \ \mathbf{(A)} \ $ thus does not even have real roots, it may be eliminated.  The balance of the choices indicate that $ \ \sec^2 x \ + \ \csc^2 x \   =   \   \sec^2 x  ·  \csc^2 x \ = \ m \ > \ 4 \ \ , $ they all seem to be acceptable.
We ought to check that the other given values for $ \ m \ $ can be attained.  Writing $$ \sec^2 \theta \ + \ \csc^2 \theta \ \ = \ \ \frac{1}{\cos^2 \theta} \ + \ \frac{1}{(1 - \cos^2 \theta)} \ \ = \ \  \frac{1}{\cos^2 \theta · (1 - \cos^2 \theta)} \ \ = \ \  m  $$
$$ \Rightarrow \ \  \cos^4 \theta \ - \   \cos^2 \theta \ + \ \frac{1}{m}  \ \ = \ \  0 \ \ , $$
we have the (presumed) solutions $ \ \cos^2 \theta \ = \ \frac12 \ \pm \ \frac{\sqrt{1 - \frac{4}{m}}}{2} \ \ . $  The solutions are real for $ \ m \ \ge \ 4 \ \ $ and, since the   term  $ \ \frac{\sqrt{1 - \frac{4}{m}}}{2}  \   $  only asymptotically approaches $ \ \frac12 \ $ as $ \ m \ \rightarrow \ \infty \ \ , $ we have valid results for $ \ \cos^2 \theta \ \ . $ All of the other equation choices have $ \ m \ > \ 4 \ \ , $ so they all have admissible real roots with $ \ \sec^2 \theta \ge 1 \ , \ \csc^2 \theta \ge 1 \ \ . $
The problem in your second method then for the choice $ \ \sec^2\theta + \csc^2\theta \ = \ p \ , \ \sec^2\theta·\csc^2\theta \ = \ q \ $ is that $ \ p = q \ \ , $ which implies that we also have
$$ \sec^2 \theta \ + \ \frac{\mathbf{p}}{\sec^2\theta} \ \ = \ \ p \ \ \Rightarrow \ \ \sec^2 \theta \ \  \overbrace{=}^{?!} \ \ p · (1 \ - \ \cos^2 \theta) \ \ , $$ which is certainly non-negative, but not always greater than or equal to $ \ 1 \ \ . $
