If $ax+b\sec(\tan^{-1}x)=c$ and $ay+b\sec(\tan^{-1}y)=c$, then ... After having struggled  yesterday with this as much as I could, I am posting this problem here:

If $ax+b\sec(\tan^{-1}x)=c$ and $ay+b\sec(\tan^{-1}y)=c$, then prove that $$\frac{x+y}{1-xy}=\frac{2ac}{a^2-c^2}$$

My attempt: Comparing both the equations, it is clear that $x=y$ (it is a dummy variable, sort of). So we basically need to find $\frac{2x}{1-x^2}$. Letting $x=\tan\theta$, the given equation becomes:
$$a \tan\theta +b \sec\theta=c$$
From here, I don't know what to do. I tried to put it in the form of $a \sin\theta -c\cos\theta=-b$, then divide and multiply by $\sqrt{(a^2+c^2)}$ to put it in the auxiliary form ($a=r\cos\theta,c=r\sin\theta$), but alas that did not help. 
By just working backwards, we see that the transformations $a=r\cos\alpha$ and $c=r\sin\alpha$ give the answer. These are also the transformations I need in order to get the auxiliary form I mentioned above, but connecting them beats me.
 A: I like the direction Brian Tung suggests; that identity was the first thing that popped out to me when I first saw this question, but you do have to do something about possible spurious solutions created by squaring, and I 'd prefer to avoid that. :) So I'll continue where you left off.
Notice that both sides of the identity you are required to prove are related to tangent sums. So let's try to work towards that direction.
$$
a \tan\theta +b \sec\theta = c\\
c \cos\theta - a \sin\theta = b$$
Let
$$a^2 + c^2 = r^2$$
So for some $\alpha$,
$$c = r \cos\alpha, \, a = r \sin\alpha$$
Also, $$\tan\alpha = a / c$$
Thus
$$\cos\alpha\cos\theta - \sin\alpha\sin\theta = \frac{b}{r}$$
$$\cos(\alpha + \theta) = \frac{b}{r}$$
$$\alpha + \theta = \pm \cos^{-1}(b/r)$$
WLOG let $\alpha + \theta$ be the positive solution, and $\alpha + \phi$ the negative solution, where $y = \tan \phi$
$$\alpha + \phi = -(\alpha + \theta)$$
 $$-2\alpha = \theta + \phi$$
Taking tangents,
$$\frac{-2\tan\alpha}{1-\tan^2\alpha} = \tan(\theta + \phi)$$
$$\frac{-2a/c}{1-a^2/c^2} = \frac{x + y}{1-xy}$$
$$\frac{2ac}{a^2-c^2} = \frac{x + y}{1-xy}$$
A: It is evident from the form of the two equations that $x$ and $y$ are both solutions for $u$ in the equation
$$
au+b \sec(\tan^{-1} u) = c
$$
We note that $\sec(\tan^{-1} u) = \sqrt{1+u^2}$, so we have
$$
au+b\sqrt{1+u^2} = c
$$
Subtract $au$ from both sides, and then square, to obtain
$$
b^2(1+u^2) = c^2-2acu+a^2u^2
$$
Collect like terms of $u$ and $u^2$ to get
$$
(a^2-b^2)u^2-2acu+c^2-b^2 = 0
$$
This is a quadratic equation with two roots, which can be equated with $x$ and $y$.  Recall that in a quadratic equation of the form $mz^2+nz+p = 0$, the sum of the roots is equal to $-n/m$, and the product of the roots is equal to $p/m$.  I presume you can take it from here.
ETA: It remains to be shown that squaring did not introduce extraneous solutions.
