# Simplify $\tan \{{1\over2}\tan^{-1}x + \tan^{-1}y\}$ .

My Process Seems Long. Here it is :

• Write $\tan^{-1}y$ as $1/2\tan^{-1}y + 1/2\tan^{-1}y$

• Add $1/2\tan^{-1}x$ to one $1/2\tan^{-1}y$

• Add result of previous to the other $1/2\tan^{-1}y$

• We will get $\tan(1/2\tan^{-1}z)$ where $$z = {{ x + 2y - xy^2 }\over{ 1 - 2xy - y^2 }}$$

• Let $\tan^{-1}z = \theta$ $\implies \tan\theta = z$

• Our expression becomes $\tan{\theta/2}$ = $\sqrt{{1-\cos\theta}\over{1+\cos\theta}}$

• Where $\cos \theta = \sqrt{1\over{1+z^2}}$

It becomes quite messy as you see. So is there a simpler way .

• Try using an identity for $\tan(a + b)$ Commented Feb 20, 2016 at 4:48
• @deinst let me tryu Commented Feb 20, 2016 at 4:54
• @deinst its simpler thanks ... Commented Feb 20, 2016 at 4:59

In most (scholar) questions like this one, the simplest way is to give names to the different $atan^{-1}(...)$. Set

$a=atan^{-1}(x) \ \leftrightarrow \ \tan(a)=x \ \ \$ and $\ \ \ b=\tan^{-1}(y) \ \leftrightarrow \ \tan(b)=y$.

$\tan(a/2+b)=\dfrac{\tan(a/2)+\tan(b)}{1-\tan(a/2)\tan(b)}=\dfrac{t+y}{1-ty} \ \ \ (1)$.

by setting $t=\tan(a/2)$.

Inverting the classical formula $\tan(a)=\dfrac{1+t^2}{1-t^2}$ gives

$t=\sqrt{\dfrac{\tan(a)-1}{\tan(a)+1}}=\sqrt{\dfrac{x-1}{x+1}} \ \ \ (2)$

The answer is obtained by plugging (2) into (1).

Let $\dfrac{\tan^{-1}x}2=A\implies x=\tan2A$ and $-\dfrac\pi2\le2A\le\dfrac\pi2$

$\implies-1\le\tan A\le1\ \ \ \ (1)$

and $\tan^{-1}y=B\implies y=\tan B$

Now $\tan\left(\dfrac{\tan^{-1}x}2+\tan^{-1}y\right)=\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

Replace the value of $\tan B$

Now $x=\tan2A=\dfrac{2\tan A}{1-\tan^2A}$

Rearrange to form a Quadratic Equation in $\tan A$

Solve for $\tan A$

Choose $\tan A$ honoring $(1)$