If $\arctan x + \arctan y = \pi/4$ show that $x+y= 1-xy$

Question

I dont really know how to start this question. So I was wondering if I could be given a hint on where to start.

If the $x+y$ in the question indicates the $\arctan x + \arctan y$ and so $1-xy$ which would also mean it would indicate $1-\arctan x\arctan y$ and then if i had divided it to the other side may lead to $\arctan(x+y)$ using compound formulae?

If what I have said is true how do I find out that $\arctan x$ is $x$ and $\arctan y$ is $y$ etc?

If it is not true, where do I begin?

Thank you!

Answer 1

Hint: If $\arctan x+\arctan y=\frac\pi4$, then $\tan(\arctan x+\arctan y)=1$.

Answer 2

According to the addition formula for the inverse tangent function we get that

$$\arctan x+\arctan y=\arctan\frac{x+y}{1-xy}$$

We know that $\arctan 1=\frac\pi4$, hence we can conclude that

$$\therefore~\frac{x+y}{1-xy}=1\Leftrightarrow x+y=1-xy$$

EDIT

As pointed out by zwim it is necessary to mention that the inverse tangent functions is injective. Otherwise we would have to restrice the values of $x$ and $y$, respectively.