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.
 A: The statement is equivalent to
$$\DeclareMathOperator{\arccot}{arccot}
\arccot7+\arccot8=\arccot3-\arccot18
$$
The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $(0,\pi/2)$ and we have to ensure that the same holds for the left-hand side, but this is easy, because $7>1$ and $8>1$, hence
$$
\arccot7+\arccot8<2\arccot1=\frac{\pi}{2}
$$
The two sides are equal if and only if their cotangents are.
Since
$$
\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}
\qquad
\cot(\alpha-\beta)=\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}
$$
this amounts to proving that
$$
\frac{7\cdot8-1}{7+8}=\frac{3\cdot18+1}{18-3}
$$
A: 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 $(a + b + c) = \text{Arccot}(3).$
A: ${tan^{-1}(1/7)+tan^{-1}(1/8)+tan^{-1}(1/18)}$

*

*${tan^{-1}  \left[  \frac{ (1/7)+(1/8) } { \left( 1 \right)  - (1/7)(1/8) }  \right]  + tan^{-1}(1/18) }$

*= ${tan^{-1}  \frac {3}{11} + tan^{-1} \frac{1}{18}}$

*=${tan^{-1} \left[  \frac {(3/11)+(1/18)} {\left (1 \right) - (3/11)(1/18)} \right] }$

*= ${tan^{-1} \frac{1}{3} }$ or ${cot^{-1} {3}}$
A: Obviously $\pi+\cot^{-1} 3 \neq \cot^{-1} 3$, since $\pi \neq 0$. Your second method is incorrect because $$\tan^{-1} a+\tan^{-1}b+\tan^{-1}c=\tan^{-1} \left(\frac {a+b+c-abc}{1-ab-bc-ac}\right)$$ is only true for $a,b,c>0$ if $ab+bc+ac<1$.
It is better for you to combine the $\arctan$s two at a time. We have:
$$\tan^{-1}7+\tan^{-1}8=\pi-\tan^{-1} \frac {3}{11} {\tag 1}$$
So $$\tan^{-1}7+\tan^{-1}8+\tan^{-1}18=\left(\pi-\tan^{-1} \frac {3}{11}\right)+\tan^{-1}18=\pi+\left(\tan^{-1}18-\tan^{-1} \frac {3}{11}\right)=\pi+\left(\tan^{-1}\frac {195}{65}\right)=\pi+\tan^{-1} 3$$
Thus we get $$\frac {3\pi}{2}-(\tan^{-1}7+\tan^{-1}8+\tan^{-1} 18)=\frac {3\pi}{2}-(\pi+\tan^{-1}3)=\frac {\pi}{2}-\tan^{-1}3=\cot^{-1}3$$ as expected.
Note:
$(1)$: I used the formula $\tan^{-1}a+\tan^{-1}b=\pi+\tan^{-1} \left( \frac {a+b}{1-ab}\right)$ which is true iff $ab>1, a>0, b>0$.
●For calculating $\tan^{-1}18-\tan^{-1}\frac {3}{11}$, I used $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac {a-b}{1+ab}\right)$ which is true iff $ab>-1$.
