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According to my book

$$\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$ when $x<0$, $y<0$, and $xy>1$.

I can't understand one thing out here that when the above mentioned conditions on $x$ and $y$ are followed then the denominator of the argument of $\tan^{-1}(1-xy)$ become negative while the numerator too becomes negative and $x$ and $y$ both are less than zero. Now as both the numerator and denominator are negative the arguments i.e $\left(\frac{x+y}{1-xy}\right)$becomes positive overall.

Now why do we add $\pi$ to the expression when we are already having a positive argument which can be found in the first quadrant which is found in principal range. Now is it because we can also find the postive tangent function in third quadrant also? If this is so, why has this been mentioned up as a separate identity rather than another solution?

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    $\begingroup$ The problem is that $\tan^{-1}x+\tan^{-1}y$ "overflows". Meaning, it is not necessarily in the usual range $(-\pi/2,\pi/2)$ of $\arctan$. See what happens when, for example, $x=y=-\sqrt3$. $\endgroup$ – Jyrki Lahtonen Jun 23 '16 at 20:01
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    $\begingroup$ For the right values of $x,y, \tan^{-1} x + \tan^{-1} y$ are less than $-\pi$ $\endgroup$ – Doug M Jun 23 '16 at 20:05
  • $\begingroup$ Indeed $${{\tan }^{-1}}x+{{\tan }^{-1}}y=\left\{ \begin{align} & \qquad\quad{{\tan }^{-1}}\frac{x+y}{1-xy}\quad,\quad xy<1 \\ & \quad\pi+{{\tan }^{-1}}\frac{x+y}{1-xy}\quad,\quad xy>1\quad,\quad x>0 \\ & -\pi+{{\tan }^{-1}}\frac{x+y}{1-xy}\quad,\quad xy>1\quad,\quad x<0 \\ \end{align} \right.$$ $\endgroup$ – Behrouz Maleki Jun 23 '16 at 20:14
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Consider $y$ constant and the functions $$ f(x)=\arctan x+\arctan y \qquad g(x)=-\pi+\arctan\frac{x+y}{1-xy} $$ Then $f'(x)=1/(1+x^2)$, whereas $$ g'(x)=\frac{1}{1+\dfrac{(x+y)^2}{(1-xy)^2}}\frac{1-xy+y(x+y)}{(1-xy)^2}= \frac{1+y^2}{1+x^2+y^2+x^2y^2}=\frac{1}{1+x^2} $$ Therefore the two functions differ by a constant in every connected component of their domain.

Suppose $x<0$, $y<0$ and $xy>1$. Then we can consider the limit at $-\infty$ of $f$ and $g$: $$ \lim_{x\to-\infty}f(x)=-\frac{\pi}{2}+\arctan y $$ while $$ \lim_{x\to-\infty}g(x)=-\pi+\arctan\frac{1}{-y}=-\pi+\frac{\pi}{2}+\arctan y $$ due to $$ \arctan y+\arctan\frac{1}{y}=-\frac{\pi}{2} $$ for $y<0$.

Thus $f(x)=g(x)$ in the stated domain.

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From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),

$$\arctan x+\arctan y=\begin{cases} \arctan\frac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\frac{x+y}{1-xy} & \mbox{if } xy>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } xy=1\end{cases} $$

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  • $\begingroup$ I don't get what Loney is trying to tell in his last statement. He says that $\gamma$ is a negative angle and $\tan(\pi + \gamma)= \tan \gamma$ but $\gamma$ is still negative :/ How does that make a difference? Could you explain through an example.. $\endgroup$ – Abcd Feb 8 '18 at 12:27
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This is not a complete answer, but rather a response to the following doubt:

Now why do we add π to the expression when we are already having a positive argument ...

Actually, we subtract $\pi.$ But how are we to understand intuitively that this is something we might want to do?

As you observed, under the given conditions on $x$ and $y,$ we know that $\frac{x+y}{1-xy} > 0$ and therefore $\tan^{-1}\left(\frac{x+y}{1-xy}\right) > 0$ as well.

But the given conditions include $x< 0$ and $y< 0$, from which it follows that $\tan^{-1} x < 0,$ that $\tan^{-1} y < 0,$ and that $\tan^{-1} x + \tan^{-1} y < 0.$

We cannot have an equation with a negative number on the left and a positive number on the right, can we? But we can add or subtract something on the left or right in order to make the two sides equal after all. What is to be shown then is that the thing to add or subtract is a constant for all $x$ and $y$ that satisfy the given conditions, and that subtracting the particular constant $\pi$ from the right side will make the equation satisfied.

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