1

Show in two different way that,

$$\arctan\left({\frac{x+y}{1-xy}}\right) = \arctan\left(x\right) + \arctan\left(y\right) $$

The first way I know derive from the addition formula of tangent

$$ \frac{ \tan\left(x\right)+ \tan\left(y\right)}{1- \tan(x) \tan(y)} = \tan\left(x+y \right). $$

Guy Fsone
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  • Now replace $x$ by $\arctan x$, $y$ by $\arctan y$, and apply $\arctan$ to both sides of the equation. –  Jan 07 '17 at 16:49
  • http://www.math-only-math.com/arctan-x-plus-arctan-y.html – Behrouz Maleki Jan 07 '17 at 16:52
  • The proposition is not an identity. Related :http://math.stackexchange.com/questions/1724348/what-is-arctanx-arctany OR http://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Jan 08 '17 at 10:41

3 Answers3

3

Differentiating, $$ \begin{align} \frac{d}{dx} \arctan{\left( \frac{x+y}{1-xy} \right)} &= \frac{1}{1+(x+y)^2/(1-xy)^2} \left( \frac{1}{1-xy} - \frac{-(x+y)y}{(1-xy)^2} \right) \\ &= \frac{ 1-xy + xy+y^2 }{(1-xy)^2 + (x+y)^2} \\ &= \frac{ 1+y^2 }{1-2xy+x^2y^2+x^2+2xy+y^2} \\ &= \frac{ 1+y^2 }{1+x^2+y^2+x^2 y^2} \\ &= \frac{ 1+y^2 }{(1+y^2)(1+x^2)} = \frac{1}{1+x^2}, \end{align} $$ and then integrating gives $$ \arctan{\left( \frac{x+y}{1-xy} \right)} = \arctan{x}+C(y). $$ Putting $x=0$ gives $C(y)=\arctan{y}$.

Chappers
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1

There's always a certain "crystal ball" aspect to a question such as this, but another approach is to show that for each (fixed) real number $y$, the function $$ f(x) = \arctan\left(\frac{x + y}{1 - xy}\right) - \arctan x - \arctan y $$ has vanishing derivative (throughout its domain), and vanishes at some point (such as where $y = -x$).

(Strictly speaking, of course, the two are not equal, since the left-hand side is undefined where $xy = 1$ and the right-hand side is defined for all real $x$ and $y$.)

Andrew D. Hwang
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0

$$(1+ix)(1+iy)=(1-xy)+i(x+y)$$

$$\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$$

$$\arg(a+bi)=\arctan \frac{b}{a}$$

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

Remarks:

  1. Depending on your perspective, this might be more of a proof of the 2nd line.
  2. The third line is not entirely true, as it can require some adjustment depending on the signs of $a,b$
  3. This is also true of the final result, actually.
πr8
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