## Homework Statement:

Prove the following identity . If $f(x) = \arctan(x) + \arctan(y)$ then prove that
$$f(x) = \begin{cases} \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \qquad xy < 1 \\ \pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x>0, y > 0, xy > 1 \\ -\pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x < 0, y < 0, xy > 1 \end{cases}$$

## Relevant Equations:

Equations of inverse tangent functions
Now following is my attempt for the very first one. Let $\arctan(x) = \alpha$ and $\arctan(y) = \beta$. So, I have $x = \tan(\alpha)$ and $y = \tan(\beta)$. Since the domain is restricted to define inverse function, I have that $\alpha \in (-\pi/2, \pi/2)$ and $\beta \in (-\pi/2, \pi/2)$. Now, I use the following identity for addition of angles

$$\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta) }{ 1- \tan(\alpha) \tan(\beta) }$$

On the left hand side, we have $\alpha + \beta$ as the argument for the $tan$ function. But since the domain of the tan function is restricted, we must have that $\alpha + \beta \in (-\pi/2, \pi/2)$. This further restricts the values which $\alpha$ and $\beta$ can take. So, we have $\alpha \in (-\pi/4, \pi/4)$ and $\beta \in (-\pi/4, \pi/4)$. This means the following

$$- \frac{\pi}{4} < \alpha < \frac{\pi}{4}$$
$$- \frac{\pi}{4} < \beta< \frac{\pi}{4}$$

Now, I use the fact that $\tan$ is an increasing function. So, I get $- 1 < \tan(\alpha) < 1$ and $-1 < \tan(\beta) < 1$. Which means that $x \in (-1,1)$ and $y \in (-1,1)$. From this, its easy to prove that $xy < 1$ and plugging for x and y, we get the required first identity.

$$\alpha + \beta = \arctan \Big( \frac{x+y}{1 - xy} \Big)$$
$$\arctan(x) + \arctan(y) = \arctan \Big( \frac{x+y}{1 - xy} \Big)$$

Is my reasoning right ? For the other two identities, I have no clue how to proceed.
Thanks

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PeroK
Homework Helper
Gold Member
Homework Statement:: Prove the following identity . If $f(x) = \arctan(x) + \arctan(y)$ then prove that
$$f(x) = \begin{cases} \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \qquad xy < 1 \\ \pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x>0, y > 0, xy > 1 \\ -\pi + \arctan\bigg( \frac{x+y}{1 - xy} \bigg), \quad x < 0, y < 0, xy > 1 \end{cases}$$
Relevant Equations:: Equations of inverse tangent functions

Now following is my attempt for the very first one. Let $\arctan(x) = \alpha$ and $\arctan(y) = \beta$. So, I have $x = \tan(\alpha)$ and $y = \tan(\beta)$. Since the domain is restricted to define inverse function, I have that $\alpha \in (-\pi/2, \pi/2)$ and $\beta \in (-\pi/2, \pi/2)$. Now, I use the following identity for addition of angles

$$\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta) }{ 1- \tan(\alpha) \tan(\beta) }$$

On the left hand side, we have $\alpha + \beta$ as the argument for the $tan$ function. But since the domain of the tan function is restricted, we must have that $\alpha + \beta \in (-\pi/2, \pi/2)$. This further restricts the values which $\alpha$ and $\beta$ can take. So, we have $\alpha \in (-\pi/4, \pi/4)$ and $\beta \in (-\pi/4, \pi/4)$. This means the following

$$- \frac{\pi}{4} < \alpha < \frac{\pi}{4}$$
$$- \frac{\pi}{4} < \beta< \frac{\pi}{4}$$

Now, I use the fact that $\tan$ is an increasing function. So, I get $- 1 < \tan(\alpha) < 1$ and $-1 < \tan(\beta) < 1$. Which means that $x \in (-1,1)$ and $y \in (-1,1)$. From this, its easy to prove that $xy < 1$ and plugging for x and y, we get the required first identity.

$$\alpha + \beta = \arctan \Big( \frac{x+y}{1 - xy} \Big)$$
$$\arctan(x) + \arctan(y) = \arctan \Big( \frac{x+y}{1 - xy} \Big)$$

Is my reasoning right ? For the other two identities, I have no clue how to proceed.
Thanks
I don't see where you used $xy < 1$ here.

Do you understand why $f(x)$ is not always equal to $g(x) = \arctan(\frac{x + y}{1 - xy})$, depending on $x$ and $y$?

Well $x$ and $y$ are in $\mathbb{R}$ and since I reasoned that $x, y \in (-1,1)$, we can take various cases of $x$ and $y$ and then prove that $xy < 1$. I thought we are supposed to prove this.

PeroK
Well $x$ and $y$ are in $\mathbb{R}$ and since I reasoned that $x, y \in (-1,1)$, we can take various cases of $x$ and $y$ and then prove that $xy < 1$. I thought we are supposed to prove this.
If $xy < 1$, you could have $x = \frac 1 3, \$y = 2$. What about my main question: Do you understand why$f(x)$is not always equal to$g(x) = \arctan(\frac{x + y}{1 - xy})$, depending on$x$and$y$? No, I am confused about the question. PeroK Science Advisor Homework Helper Gold Member No, I am confused about the question. Try some numbers. See what you get. Ok, so the domain of$\arctan$should be$(-\pi/2, \pi/2)$, right ? I will try some values PeroK Science Advisor Homework Helper Gold Member Ok, so the domain of$\arctan$should be$(-\pi/2, \pi/2)$, right ? I will try some values The domain of$\arctan$is$\mathbb R$. The range is$(-\pi/2, \pi/2)$. It's a tricky question. I tried few values of$x,y$so that$xy < 1$on WolframAlpha and yes in this case, we have that  \arctan(x) + \arctan(y) = \arctan\Big( \frac{x+y}{1-xy} \Big)  So, how do I proceed proving this ? How would my argument change ? PeroK Science Advisor Homework Helper Gold Member I tried few values of$x,y$so that$xy < 1$on WolframAlpha and yes in this case, we have that  \arctan(x) + \arctan(y) = \arctan\Big( \frac{x+y}{1-xy} \Big)  So, how do I proceed proving this ? How would my argument change ? What I meant was to try to find values of$x, y$where this is not true and see why. The confusing thing, surely, is why that equation is not true for all$x, y$. Why do you sometimes have to add$\pm \pi$? So, for$x=10$and$y = 1/3$, I get$\arctan(x) + \arctan(y) = 1.793$, which is greater than$\pi/2$, so this is is beyond the range of arctan function and$\arctan((x+y)/(1-xy)) = -1.35$and this is within the range of arctan function. PeroK Science Advisor Homework Helper Gold Member So, for$x=10$and$y = 1/3$, I get$\arctan(x) + \arctan(y) = 1.793$, which is greater than$\pi/2$, so this is is beyond the range of arctan function and$\arctan((x+y)/(1-xy)) = -1.35$and this is within the range of arctan function. ... and the difference is$1.793 - (-1.35) = 3.14 = \pi$. What I would do first is draw the graph of$\tan$and show various possibilities for$\alpha, \beta$and$\alpha + \beta$. Because of the nature of the$\arctan$function, this isn't going to come out purely algebraically. You're going to need the graph of$\tan$. So, is there no algebraic proof of this ? PeroK Science Advisor Homework Helper Gold Member So, is there no algebraic proof of this ? I said "purely" algebraically. The problem is knowing when to add$\pm \pi$. That's not going to come out very easily without using the graph. Ok. Thanks If math analysis is allowed then one can consider derivative$\frac{\partial f}{\partial x}$haruspex Science Advisor Homework Helper Gold Member is there no algebraic proof of this ? The difficulty is that the arctan function is discontinuous. You know that$\arctan(\frac{x+y}{1-xy})$will produce the right value modulo$\pi$. It remains to show that for the second case it produces a value in the range$(-3\pi/2, -\pi/2)$, etc. If math analysis is allowed then one can consider derivative$\frac{\partial f}{\partial x}$How does that help with the question asked by the OP, namely, dealing with the different ranges? How does that help with the question asked by the OP, namely, dealing with the different ranges? Let us regard$y$as a parameter. We have two functions f(x)=\arctan x+\arctan y,\quad g(x)=\arctan\Big(\frac{x+y}{1-xy}\Big). Assume for example that$y>0,\quad x>1/y$. We obviously have$g'(x)=f'(x)$. And these derivatives are continuous for$x>1/y$. Thus for such$x$it follows that$g(x)=f(x)+const.$The constant may depend on the parameter$y$. Passing to a limit as$x\to 1/y+$we get -\pi/2=\arctan(1/y)+\arctan y+const. On the other hand \frac{d}{dy}(\arctan(1/y)+\arctan y)=0 and thus (chek for$y=1##) we get
$$\arctan(1/y)+\arctan y=\pi/2\Longrightarrow const=-\pi$$