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?
 A: 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 $\tan^2(\pi /4 +A)$.
A: Let $y=\dfrac{\tan(x+A)}{\tan(x-A)}$  and observe that $\tan^2\left(\dfrac\pi4+A\right)=\dfrac{1-\cos2\left(\dfrac\pi4+A\right)}{1+\cos2\left(\dfrac\pi4+A\right)}=\dfrac{1+\sin2A}{1-\sin2A}$
$$\implies\dfrac{y+1}{y-1}=\dfrac{\sin(x+A+x-A)}{\sin(x+A-(x-A))}$$
$$\left(\dfrac{y+1}{y-1}\right)^2=\dfrac{\sin^22x}{\sin^22A}\le\dfrac1{\sin^22A}$$
$$\iff y^2\cos^22A-2y(1+\sin^22A)+\cos^22A\ge0$$
Now we know that if $(t-a)(t-b)\ge0$ 
either $t\ge$max$(a,b)$ or $t\le$min$(a,b)$  i.e., $t$ cannot lie between $a,b$
Can you identify $a,b$ here?
