How to find cotangent?

Question

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 $-\sqrt{5}/5$

Answer 1

Since by Vieta's formulas one has $$\tan x+\tan y=-\frac{-3\sqrt 5}{1}=3\sqrt 5,\ \ \ \tan x\tan y=\frac{2}{1}=2,$$ one has $$3\cot(x+y)=3\cdot\frac{1}{\tan(x+y)}=3\cdot\frac{1-\tan x\tan y}{\tan x+\tan y}=\frac{3(1-2)}{3\sqrt 5}=-\frac{\sqrt 5}{5}.$$

Answer 2

\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, $$ 3\cot(x+y) = \frac{-1}{\sqrt 5} = \frac{-\sqrt 5}{5}. $$

Answer 3

From quadratic equation given,

$$\tan{x}+\tan{y}=3\sqrt5$$ $$\tan{x}\tan{y}=2$$

So by the Compound Angle Formula,

$$\tan{(x+y)}=\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}=\frac{3\sqrt5}{-1}=-3\sqrt5$$

So we have

$$3\cot{(x+y)}=3\cdot{\frac1{\tan{(x+y)}}}=3\cdot{\frac1{-3\sqrt5}}=\frac{-\sqrt5}{5}$$