# To prove: cot <mrow class="MJX-TeXAtom-ORD"> &#x2212;<!-- − --> 1

To prove: ${\mathrm{cot}}^{-1}7+{\mathrm{cot}}^{-1}8+{\mathrm{cot}}^{-1}18={\mathrm{cot}}^{-1}3$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Blaine Andrews
The statement is equivalent to
$\mathrm{arccot}7+\mathrm{arccot}8=\mathrm{arccot}3-\mathrm{arccot}18$
The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $\left(0,\pi /2\right)$ 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
$\mathrm{arccot}7+\mathrm{arccot}8<2\mathrm{arccot}1=\frac{\pi }{2}$
The two sides are equal if and only if their cotangents are.
Since
$\mathrm{cot}\left(\alpha +\beta \right)=\frac{\mathrm{cot}\alpha \mathrm{cot}\beta -1}{\mathrm{cot}\alpha +\mathrm{cot}\beta }\phantom{\rule{2em}{0ex}}\mathrm{cot}\left(\alpha -\beta \right)=\frac{\mathrm{cot}\alpha \mathrm{cot}\beta +1}{\mathrm{cot}\beta -\mathrm{cot}\alpha }$
this amounts to proving that
$\frac{7\cdot 8-1}{7+8}=\frac{3\cdot 18+1}{18-3}$