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 ?

You started by noting $\cot^{-1}21=\tan^{-1}\frac{1}{21}$ etc. The problem came when you encountered a negative argument. Since all the functions of interest are odd, $\cot^{-1}(-8)=-\cot^{-1}8=-\tan^{-1}\frac{1}{8}$, no $\pi$ involved. (Another way to prove $\tan^{-1}\frac{1}{21}+\tan^{-1}\frac{1}{13}=\tan^{-1}\frac{1}{8}$ is to use the identity $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}$, which for $a=1/m,\,b=1/n$ simplifies to $\tan^{-1}\frac{m+n}{mn-1}$.)
