In solving an inverse trigonometric equation is it sufficient to check for one case for the existence of a solution? 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



 A: 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.
