Show value of $\frac{\tan(x+A)}{\tan(x-A)}$ can not lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$ Let $$y=\frac{\tan(x+A)}{\tan(x-A)}.$$
Show that $y$ can't lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$.
I tried by contradiction. Assumed $$\tan^2\left(\frac{\pi}{4} -A\right)<y$$
and the counterpart's expression. Got a quadratic inequality in $\tan x$ from the former.
Solved and obtained a range of $\tan x$. (A large expression). Do I have to put the range of $\tan x$ in $y$, or is there any other method?

Use the short hands $t=\tan x$, $a=\tan A$ to rewrite $y=\frac{\tan(x+A)}{\tan(x-A)}$ as a quadratic equation in $t^2$ as you attempted,
$$(1+y)at^2+(1+a)(1-y)t+a(1+y)=0$$
Considering that the range over which $y$ can't  lie corresponds to no real solutions for $t$, which requires that the discriminant is negative, i.e.
$$(1+a^2)^2(1-y)^2-4a^2(1+y)^2<0$$
Reexpress the inequalities as 
$$ -\frac{2|a|}{1+a^2} < \frac{1-y}{1+y} < \frac{2|a|}{1+a^2} $$
Rearrange to get
$$ \left( \frac{1-|a|}{1+|a|} \right)^2 < y <  \left( \frac{1+|a|}{1-|a|} \right)^2 \tag 1$$
Now, examine two cases of $A$:
Case 1) A in the 1st and 3rd quadrants. Then, $|a| = \tan A$ and the inequalities becomes
$$ \left( \frac{1-\tan A}{1+\tan A} \right)^2 < y <  \left( \frac{1+\tan A}{1-\tan A} \right)^2$$
which is 
$$ \tan^2 (\pi /4-A)< y < \tan^2 (\pi /4+A)$$
Case 2) A in the 2nd and 4th quadrants. Then, $|a| = -\tan A$ and the inequalities becomes
$$ \tan^2 (\pi /4+A)< y < \tan^2 (\pi /4-A)$$
Thus, for any value of $A$, $y$ can't lie between $\tan^2(\pi /4 -A)$ and $\tan^2(\pi /4 +A)$.
