<s>[INST] Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$ 
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$

My Attempt
\begin{align}
\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{4-3}{1+4.3}+\tan^{-1}\frac{5-4}{1+5.4}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}5-\tan^{-1}3\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{8}=\pi
\end{align}
My reference gives the solution $0$, so what's going wrong with my attempt ?
 [/INST] Hint.
$$
\tan(\mbox{arccot}(a)) = \frac 1a
$$
and
$$
\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}
$$
so
$$
\tan(\mbox{arccot}(a)+\mbox{arccot}(b)) = \frac{a+b}{a b -1}
$$
etc.
NOTE
You can try now 
$$
\tan(\mbox{arccot}(a)+\mbox{arccot}(b)+\mbox{arccot}(c)) = d = \frac{a (b+c)+b c-1}{abc-(a+b+c)}
$$
and then
$$
\mbox{arccot}(a)+\mbox{arccot}(b)+\mbox{arccot}(c) = \mbox{arctan}(d)
$$
after substitution we have $d=0$ and $\arctan(0) = 0$
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