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?

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.
