# Inverse trigonometry problem

## 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}##.

## 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
Homework Helper
Gold Member
Do you know a relationship between tan and sec?

Delta2
Homework Helper
Gold Member
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

Last edited:
@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!

• Delta2
haruspex
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