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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.

PM 2Ring
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GRrocks
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  • You can't just assume a=rcosθ and c=rsinθ, it's a little subtler than that. But you can use a different angle parameter and see where that leads; see below. – PM 2Ring Jun 02 '16 at 08:30
  • @PM2Ring I know I did that by mistake :P....I did mention 'auxiliary' angle so I meant the same thing..thanks for pointing that out anyways – GRrocks Jun 02 '16 at 08:32
  • No worries. You should probably [edit] your question to make it clear that the parameter in that pair of equations for $a$ and $c$ isn't the $\theta$ of your earlier equations. – PM 2Ring Jun 02 '16 at 08:49

2 Answers2

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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, $$\frac{-2\tan\alpha}{1-\tan^2\alpha} = \tan(\theta + \phi)$$ $$\frac{-2a/c}{1-a^2/c^2} = \frac{x + y}{1-xy}$$ $$\frac{2ac}{a^2-c^2} = \frac{x + y}{1-xy}$$

PM 2Ring
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2

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.

Brian Tung
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