<s>[INST] To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{18}$ = $(\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8})+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\frac{1}{8}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{15}{56}}{\frac{55}{56}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{3}{11} +\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\frac{1}{18}}$ = $\tan^{-1}\frac{\frac{65}{198}}{\frac{195}{198}}$ = $\tan^{-1}\frac{1}{3}$ = $\cot^{-1}3$
Second Method: we know that $\tan^{-1}x+\cot^{-1}x = \frac{π}{2}$ for $x \in \Bbb R$. So $\cot^{-1}x = \frac{π}{2} - \tan^{-1}x$.
Now $$\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \frac{π}{2} - \tan^{-1}7+ \frac{π}{2} - \tan^{-1}8+ \frac{π}{2} - \tan^{-1}18 \implies \\ \frac{3π}{2} - (\tan^{-1}7+\tan^{-1}8+\tan^{-1}18) = \frac{3π}{2} - \tan^{-1}\frac{7+8+18-7×8×18}{1-7×8-8×18-7×18} \\ = \frac{3π}{2} - \tan^{-1}\frac{-975}{-325} = \frac{3π}{2} - \tan^{-1}3 = π + (\frac{π}{2} - \tan^{-1}3) = π + \cot^{-1}3 .$$ Also we know that $\tan x$ and $\cot x$ are periodic function with period $π$. Please help me in Second Method. Is $π + \cot^{-1}3 = \cot^{-1}3$ ?. If yes, then elaborate it.
 [/INST] I can't resist, even though your request for a critique makes this response somewhat off-topic.  Let the 3 LHS angles be denoted as $a,b,c$, and use the formula
$$\cot(a + b) = \frac{[\cot(a)\cot(b)] - 1}{\cot(a) + \cot(b)}.$$
This gives $\cot(a + b) = \frac{55}{15} = \frac{11}{3}.$
Therefore,
$$\cot[(a+b) + c] = \frac{\left[\left(\frac{11}{3}\right)\left(18\right)\right] - 1}{\frac{11}{3} + 18} = 3.$$
Further, since each of $a,b,c$ are in the first quadrant, as is $(a + b + c)$, you have that $(a + b + c) = \text{Arccot}(3).$
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