This has been bugging me for quite a while.
$\arctan x + \arctan y$ is defined as $$f(x) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\[1.5ex]
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\[1.5ex]
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$ Any book that i've picked up solves the inverse trig equation only for one case.
An example
They've proceeded to solve the equation by assuming $\frac{x+1}{x-1}\times \frac{x-1}{x} < 1$ $\color{#32CD32 }{(xy<1)}$ and after solving the equation the obtained value of x = 2 does not satisfy the aforementioned condition they've simply said that there is no solution to the given equation.
Why not solve it for other assumptions $x>0, y>0 , xy>1$ and $x<0,y<0, xy>1$ in other words how is finding the solution for one of the cases sufficient for us to declare that no solution exists?
Solution of the taken example
Best Answer
You are completely justified in viewing that solution with suspicion. The explanation is inadequately stated.
There are multiple ways to approach this. But first, let me digress by pointing out a misapprehension in your post: you say that "$\arctan x + \arctan y$ is defined by"
$$\arctan x + \arctan y = \begin{cases} \arctan\left(\frac{x+y}{1-xy}\right) & xy < 1\\[1.5ex] \pi + \arctan\left(\frac{x+y}{1-xy}\right) & x > 0, y > 0, xy > 1\\[1.5ex] -\pi + \arctan\left(\frac{x+y}{1-xy}\right) & x < 0, y < 0, xy > 1\end{cases}$$
This is not a definition. The definition of $\arctan x + \arctan y$ is simply to take the arctangent of $x$ and the arctangent of $y$ and add them together. This case-by-case expression is just a formula derived from that definition.
The arctangent takes on values between $-\frac \pi 2$ and $\frac \pi 2$, so the sum of two arctangents can range anywhere between $-\pi$ to $\pi$. Note that $$\arctan\left(\frac{x+y}{1-xy}\right)$$ gives values between $-\frac \pi 2$ and $\frac \pi 2$, while $$\pi + \arctan\left(\frac{x+y}{1-xy}\right)$$ gives values between $\frac \pi 2$ and $\pi$,
$$-\pi + \arctan\left(\frac{x+y}{1-xy}\right)$$ gives values between $-\pi$ and $-\frac \pi 2$ (under the restrictions on $x$ and $y$).
Because the latter two expressions give values outside of $-\frac \pi 2$ to $\frac \pi 2$, they can never be equal to the arctangent of a single number.
This is why only the first expression needed to be checked: The sum of the two tangents in the equation is supposed to be $-\tan^{-1} 7 = \tan^{-1}(-7)$, and so the sum cannot be either of the other two formulas.
However, the solution should have mentioned this, instead of just using the first expression without explanation. But before I can actually say that the book was wrong, I'd have to examine the context, to see if they had already made mention of this restriction before the example.