Inversing the cotangent

First, consider the following. For complex numbers $z_1, z_2,\dots,z_n$ :

$arg(z_1z_2\dots z_n) = arg(z_1)+arg(z_2)+\dots+arg(z_n)$

Where $z=a+bi$ , $arg(z) = cot^{-1}(\frac{a}{b})$

Therefore we have the following formula:

$\cot\bigg(\displaystyle\sum_{k=1}^n cot^{-1}(z_k)\bigg)=\frac{Re[p]}{Im[p]}$ where $p=z_1z_2\dots z_n$

In this case, $p=(3+i)(7+i)(13+i)(21+i)=5100+3400i$

$Re[p]=5100$ , $Im[p]=3400$ , and $\frac{Re[p]}{Im[p]}=\frac{3}{2}$ .

Plugging this into our problem, our answer becomes $10\bigg(\frac{3}{2}\bigg)=\boxed{15}$

cot^(-1)(x)+cot^(-1)(y)=cot^(-1)[(xy-1)/x+y] Using the above formula we get cot^(-1)(3)+cot^(-1)(7)=cot^(-1)(2) and cot^(-1)(13)+cot^(-1)(21)=cot^(-1)(8) Again cot^(-1)(2)+cot^(-1)(8)=cot^(-1)(3/2). Now 10cot[cot^(-1)(3/2)]=10*3/2=15(Answer)

Stop being unoriginal and copying AIME problems! Seriously, its really annoying!

$10 cot [ cot^{-1}(3) + cot^{-1}(7) + cot^{-1}(13) + cot^{-1}(21) ]$ $10 cot [ tan^{-1}{\frac13} + tan^{-1}{\frac17} + tan^{-1}{\frac1{13}} + tan^{-1}{\frac1{21}} ]$ $10 cot [ tan^{-1}{\frac12} + tan^{-1}{\frac18}]$ $10 cot [ tan^{-1}{\frac{10}{15} ] = 10 cot [ cot^{-1}{\frac{15}{10}}} ]$ $15$

Formulas used above :

$tan^{-1} x + tan^{-1} y = tan^{-1} \frac{x+y}{1-xy}$

$cot^{-1} x = tan^{-1} \frac 1x$ and vice-versa