How to derive the formula for $\arctan(x) + \arctan(y)$ depending on $x,y$?

Question

I was trying to derive the following formula $$\arctan(x)+\arctan(y) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\[1.5ex] \pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\[1.5ex] -\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$

I proceeded this way

$$\tan{(A+B)}= \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$$ $$\arctan(\tan{(A+B)})=\arctan\bigg(\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\bigg)$$ $$A+B=\arctan\bigg(\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\bigg)$$ $$\tag*{$\frac{-\pi}{2}<A+B<\frac{\pi}{2}$}$$ $$A=\arctan(\tan A)$$ $$\tag*{$\frac{-\pi}{2}<A<\frac{\pi}{2}$}$$ $$B=\arctan(\tan B)$$ $$\tag*{$\frac{-\pi}{2}<B<\frac{\pi}{2}$}$$ $$\arctan(\tan A) + \arctan(\tan B)=\arctan\bigg(\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\bigg)$$ $$\tan A=x$$ $$\tan B=y$$ $$\arctan(x) + \arctan(y)=\arctan\bigg(\frac{x+y}{1-xy}\bigg)$$

Now from here ownwards I don’t know how It gets converted to 3 different definitions. You’re help will be highly appreciated.

Answer 1

Let $y\in\Bbb R$ and let $f(x)=\arctan\left(\frac{x+y}{1-xy}\right)$. Then, by the chain rule,\begin{align}f'(x)&=\frac{\frac{y^2+1}{(1-xy)^2}}{1+\left(\frac{x+y}{1-xy}\right)^2}\\&=\frac{y^2+1}{(1-xy)^2+(x+y)^2}\\&=\frac{y^2+1}{1+x^2y^2+x^2+y^2}\\&=\frac{y^2+1}{(1+y^2)(1+x^2)}\\&=\frac1{1+x^2}\\&=\arctan'(x).\end{align}So, $f-\arctan$ is constant. But $f(0)-\arctan(0)=\arctan(y)$.

Answer 2

If you think of the graph of $tan(x)$ or $arctan(x)$, then you see that we need to be careful about our domains when saying things like $x=tan\left( arctan\left( x\right) \right)$ and so on. Not only do we need to be careful, but we need to keep track at each stage what exactly is true for each possible values of $x$ and $y$. This is important in general, but especially so in periodic and discontinuous functions such as $tan$.

As the calculations are consistent we can analyse the formula after the fact as follows:

• When $xy<1$, then we have that $\left( \dfrac{x+y}{1-xy}\right)$ has positive denominator, and so in particular is not zero.

• When $xy>1$, then $x$ and $y$ have the same sign so they are either both negative or both positive. This also means $\left( \dfrac{x+y}{1-xy}\right)$ has the opposite sign as $x,y$.

• If they are both positive, then since we assume $x,y$ to be primitive, their arctans must be negative and so is the sum of these arctans. However, $\left( \dfrac{x+y}{1-xy}\right)$ has negative sign so it’s $arctan$ is negative. Since we are adding just two values, this can only have jumped by a single $\pi$-period and so we add $\pi$ to bring the solution into the correct period.

• Similarly, if they are both negative then we have a negative sum which disagrees with our now positive fraction having positive $arctan$, but this jump cannot be more than one $\pi$-interval so we subtract $\pi$.

I hope this helps! Stay safe

Answer 3

Note $\tan \theta = t\implies \theta =k\pi +\arctan t$. So

$$\tan(\arctan x+\arctan y)=\frac{x+y}{1-xy}\tag1$$

yields different inverses for $\arctan x+\arctan y$, depending on the three possibilities for its range.

Range 1: $x,y>0$, $xy>1\implies y>\frac1{x}$

$$\pi > \arctan x+\arctan y > \arctan x+\arctan \frac1{x} = \frac\pi2$$

which falls in the range of $\pi+\arctan()$ and (1) yields

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

Range 2: $x,y<0$, $xy>1\implies y<\frac1{x}$

$$-\pi < \arctan x+\arctan y < \arctan x+\arctan \frac1{x} = -\frac\pi2$$

which falls in the range of $-\pi+\arctan()$ and (1) yields

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

Range 3: $xy<1$

If $x$ and $y$ have the same sign, $|y| < \frac1{|x|}$ and

$$\hspace{-0.5cm}|\arctan x+\arctan y | < \arctan |x|+\arctan \frac1{|x|} = \frac\pi2$$

If $x$ and $y$ have opposite signs,

$$\hspace{-0cm}|\arctan x+\arctan y | \le \max( |\arctan x|,| \arctan y|)< \frac\pi2$$

Either inequality coincides the range of $\arctan()$ and (1) yields

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

Answer 4

Hint: Apply $\tan$ to the both sides of the equality and use the fact that $\tan$ is injective.

$\tan(\arctan(x)+\arctan(y))={{x+y}\over{1-xy}}$.