Inverse trigonometry problem

[cr7einstein]

Homework Statement

After having struggled yesterday with this as much as I could, I am posting this problem here-
if $ax+bsec(tan^{-1}x)=c$ and $ay+bsec(tan^{-1}y)=c$, then prove that $\frac{x+y}{1-xy}=\frac{2ac}{a^2-c^2}$.

Homework Equations

The Attempt at a Solution

My attempt- Comparing both the equations, $x=y$ clearly(it is a dummy variable, sort of). So we basically need to find $\frac{2x}{1-x^2}$. Put $x=tan\theta$; so 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 it with $\sqrt{(a^2+c^2)}$ and put it in the auxiliary form(using $a=rcos\theta,c=rsin\theta$); but alas; no help. By just working backwards, we see that the transformations $a=rcos\theta$ and $c=rsin\theta$ give the answer; and these are also the transforms I need to get my auxiliary form I mentioned above. But how do I connect them beats me. and Please help; I'd really appreciate it. Thanks in advance!

 

[haruspex]

Do you know a relationship between tan and sec?

 

[Delta2]

I doubt that it is necessarily true that x=y, that will hold only if one proves that the equation $ax+bsec(tan^{-1}x)=c$ has unique solution and I doubt a trigonometric equation has unique solution.

From what i can see from wolfram solver, the equation does not has unique solution in real numbers, at least not in all cases.

http://www.wolframalpha.com/input/?i=ax+bsec(arctan(x))=c

 

[cr7einstein]

@Delta² and @haruspex I found the answer-make the equation a quadratic with roots being x and y; then use properties of sum and products. OR, make an equation with an auxiliary angle, and then solve the resulting trig equation. Thanks for the time though!

 

[haruspex]

Ok.
Fwiw, I was hinting at using sec²=1+tan², from which sec(tan^-1(x))=√(1+x²). Maybe that's what you did.

 

[cr7einstein]

@haruspex yes that's what I did...and even going by my initial attempt as stated in the problem, I got the same result.