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Prove some identities for tangent of a sum and difference, and cotangent of a sum

by RoRi

Prove,

    \[ \tan(x-y) = \frac{\tan x - \tan y}{1+\tan x \tany} \]

for all x,y \in \mathbb{R} such that \tan x \tan y \neq -1.

We also conjecture and prove,

    \[ \tan (x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}, \]

and,

    \[ \cot (x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}. \]

First we prove the difference formula for tangent, \tan (x-y).
Proof. By definition of the tangent,

    \begin{align*} \tan (x-y) = \frac{\sin (x-y)}{\cos (x-y)} &= \frac{\sin x \cos y - \sin y \cos x}{\cos x \cos y + \sin x \sin y} &(\text{Difference formulas})\\ &= \frac{\sin x \cos y - \sin y \cos x}{\cos x \cos y(1 + \tan x \tan y)}\\ &= \frac{\tan x - \tan y}{1+\tan x \tany}. \qquad \blacksquare \end{align*}

Next, we prove the sum formula for tangent, \tan (x+y).
Proof. Again, starting with the definition of tangent,

    \begin{align*} \tan(x+y) = \frac{\sin(x+y)}{\cos (x+y)} &= \frac{\sin x \cos y + \sin y \cos x}{\cos x \cos y - \sin x \sin y} \\ &= \frac{\sin x \cos y + \sin y \cos x}{\cos x \cos y(1- \tan x \tan y)}\\ &= \frac{\tan x + \tan y}{1 - \tan x \tan y}. \qquad \blacksquare \end{align*}

Finally, the sum formula for the cotangent.
Proof. Starting with the definition of the cotangent,

    \begin{align*} \cot (x+y) = \frac{\cos (x+y)}{\sin (x+y)} &= \frac{\cos x \cos y - \sin x \sin y}{\sin x \cos y + \sin y \cos x} \\ &= \frac{\sin x \sin y(\cot x \cot y - 1)}{\sin x \cos y + \sin y \cos x} \\ &= \frac{\cot x \cot y - 1}{\cot x + cot y}. \qquad \blacksquare \end{align*}

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