Hyperbolic tangent addition

Hyperbolic tangent and cotangent are already the quotients of sinh and cosh. Hyperbolic addition places the sum of two real parameters on the right-hand branch. Why is the tanh of that sum not the sum of the two tanh values?

The sum point meets $x=1$

This picture keeps both inputs real and positive. Drag either one. The gray point is the first parameter on the right-hand branch. The green point is the sum. The segment from the origin through the green point meets the line $x=1$, and that height is $\tanh (x+y)$.

Away from the poles, for all complex $x$ and $y$,

$$\tanh (x\pm y)=\frac{\tanh x\pm \tanh y}{1\pm \tanh x\tanh y},$$

$$\coth (x\pm y)=\frac{1\pm \coth x\coth y}{\coth x\pm \coth y}.$$

Hyperbolic addition gives $\cosh (x+y)$ a plus in front of $\sinh x\sinh y$. Divide the two expansions by $\cosh x\cosh y$, and that plus lands in the denominator. Tangent addition writes a minus there, $1\mp \tan x\tan y$.

Freeze the two inputs

Set $y=x$. Away from the poles,

$$\tanh 2x=\frac{2\tanh x}{1+{\tanh }^{2}x},\coth 2x=\frac{1+{\coth }^{2}x}{2\coth x}.$$

This picture keeps the input real and positive, and keeps twice the input on the right-hand branch. Drag it. The gray point stays at the input. The green point is twice the parameter. The segment from the origin meets $x=1$ at $\tanh 2x$.

The chord from $(-1,0)$

Hyperbolic sine and cosine place $(\cosh x,\sinh x)$ on the right-hand branch. This picture keeps $x$ real. Drag it across zero. The horizontal step runs from $(-1,0)$ to $(\cosh x,0)$, so its length is $\cosh x+1$. The vertical step equals $\sinh x$. Their ratio is $\tanh (x/2)$. Hyperbolic addition puts the sign of $\sinh (x/2)$ back by hand. Here the vertical step already carries that sign.

Where the denominators are not zero,

$$\tanh \frac{x}{2}=\frac{\cosh x-1}{\sinh x}=\frac{\sinh x}{\cosh x+1},$$

$$\coth \frac{x}{2}=\frac{\sinh x}{\cosh x-1}=\frac{\cosh x+1}{\sinh x}.$$

A sum of tanh values

Away from the poles, for all complex $x$ and $y$,

$$\tanh x\pm \tanh y=\frac{\sinh (x\pm y)}{\cosh x\cosh y}.$$

One square from another

Each row is that square written through one of the other three. They rest on ${\cosh }^{2}x-{\sinh }^{2}x=1$ from hyperbolic sine and cosine.

Two real parameters add inside tanh once their sinh and cosh expansions are divided by $\cosh x\cosh y$. Freezing them doubles the parameter. The half-angle is the slope of the chord from $(-1,0)$. How a height between $-1$ and $1$ returns an angle is the question the inverse trigonometric functions answer.