Find $x$ such that $\arctan(3/2)+\dots=\arctan x$ Find $x$ such that
$$\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3) = \arctan x.$$

\begin{align}
& {} \quad \tan(a+b+c+d) \\
& = \frac{\tan a+\tan b+\tan c+\tan d\  \overbrace{ - \tan a\tan b\tan c - \cdots}^\text{4 terms}}{1-\  \underbrace{\tan a \tan b- \cdots}_\text{6 terms} +\tan a\tan b\tan c\tan d}
\end{align}
Therefore
\begin{align}
& {} \quad \tan\Big(\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3)\Big) \\
& = \frac{\frac32 + \frac54 + \frac{-5}2 +\frac{-8}3 - (\text{sum of four terms, each a product of three numbers})}{1 - (\text{sum of six terms, each a product of two numbers}) + (\text{a product of four numbers}))}
\end{align}
You get a number.  Then you say $\tan x=\text{that number}$, and find the tangent.
