Find $x$ such that $\arctan(3/2)+\dots=\arctan x$ Find $x$ such that
$$\arctan(3/2) + \arctan(5/4) + \arctan(-5/2) + \arctan(-8/3) = \arctan x.$$
 A: Hint: A repetition of applying the formula
$$ \tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} $$
gives you the answer.
A: You may use the fact if $x > 0$, the argument of the complex number $x+iy$ is $\tan^{-1} \left(\frac{x}{y}\right)$. So if you compute the product
$$(2+3i)(4+5i)(2-5i)(3-8i) = 920 - 531i$$
you get that 
$$\tan^{-1}\left(\frac{3}{2}\right) + \tan^{-1}\left(\frac{5}{4}\right) + \tan^{-1}\left(-\frac{5}{2}\right) + \tan^{-1}\left(-\frac{8}{3}\right) \equiv \tan^{-1}\left(-\frac{531}{920}\right) \mod 2\pi$$
By computing both side on a calculator, you can check this actually an equality.
A: Take $\tan$ of both sides, and repeatedly apply the compound angle formula.
A: Let $\omega_1= 2+3i, \omega_2=4+5i, \omega_3=2-5i , \omega_4=3-8i$. Then 
$$\arg(\omega_1)= \arctan(\frac{3}{2}), ...$$
Finc $\omega_1 \omega_2 \omega_3 \omega_4 =a+bi$ and then
$$\frac{b}{a}= \arg(\omega_1 \omega_2 \omega_3 \omega_4 ) =\arctan(x) \,.$$
A: \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.
