How to find cotangent? Need to find a $3\cot(x+y)$ if $\tan(x)$ and $\tan(y)$ are the solutions of $x^2-3\sqrt{5}\,x +2 = 0$.
I tried to solve this and got $3\sqrt{5}\cdot1/2$, but the answer is $-\sqrt{5}/5$

\begin{align}
& x^2 - 3\sqrt{5}\,x+ 2 = (x-a)(x-b) \\[6pt]
= {} & x^2 - (a+b) x + ab.
\end{align}
Therefore $3\sqrt 5= a+b$ and $2=ab$.
Hence $3\sqrt 5 = \tan x + \tan y$ and $2 = \tan x\tan y$.
So
$$
\tan(x+y) = \frac{\tan x+\tan y}{1-\tan x\tan y} = \frac{a+b}{1-ab} = \frac{3\sqrt 5}{1-2},
$$
and finally,
$$
3\cot(x+y) = \frac{-1}{\sqrt 5} = \frac{-\sqrt 5}{5}.
$$
