Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$

Question

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 ?

Answer 1

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 $\tan^{-1}\frac{m+n}{mn-1}$.)

Answer 2

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 $\arctan(0) = 0$