 #1
 904
 17
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
$$ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta) }{ 1 \tan(\alpha) \tan(\beta) } $$
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