Question 1201910
<pre>
{{{cot(arccos(-(sqrt(6))/3)^"")}}}

First look only at 

{{{arccos(-(sqrt(6))/3)}}}

That asks: What angle between 0 and 360 has {{{-(sqrt(6))/3}}} for its cosine?

It's a NEGATIVE cosine and the cosine is negative only in QII and QIII.

So it's got to be one of these two:

{{{drawing(200,200,-4,4,-4,4, line(-5,0,5,0),line(0,-5,0,5),
triangle(0,0,-sqrt(6),0,-sqrt(6),sqrt(3)),
red(arc(0,0,2,-2,0,145)),
locate(-2.3,-.1,-sqrt(6)),locate(-1.3,1.5,3)

 )}}}   {{{drawing(200,200,-4,4,-4,4, line(-5,0,5,0),line(0,-5,0,5),
triangle(0,0,-sqrt(6),0,-sqrt(6),-sqrt(3)),
red(arc(0,0,2,-2,0,216)),
locate(-2.3,.83,-sqrt(6)),locate(-1.3,-.9,3)

 )}}}

The first one is the PRINCIPLE value, since it has the lesser absolute
value. (The shorter red arc). So we calculate the other side of the triangle 
for the first one, using the Pythagorean theorem:

{{{matrix(5,5,

r^2,""="",x^2,""+"",y^2,
3^2,""="",(-sqrt(6))^2,""+"",y^2,
9,""="",6,""+"",y^2,
3,""="",y^2,"","",
"" +- sqrt(3),""="",y,"","")}}}

We choose the positive sign, since it goes up.

{{{drawing(200,200,-4,4,-4,4, line(-5,0,5,0),line(0,-5,0,5),
triangle(0,0,-sqrt(6),0,-sqrt(6),sqrt(3)),
red(arc(0,0,2,-2,0,145)),
locate(-2.3,-.1,-sqrt(6)),locate(-1.3,1.5,3),
locate(-3.1,1.2,sqrt(3))
 )}}}

That's the picture of

{{{arccos(-(sqrt(6))/3)}}}

And we want 

{{{cot(arccos(-(sqrt(6))/3)^"")}}}

Since we know that {{{cotangent=adjacent/opposite}}},

the answer is

{{{(-sqrt(6))/(sqrt(3))}}}{{{""=""}}}{{{-sqrt(6/3)}}}{{{""=""}}}{{{-sqrt(2)}}}.

Edwin</pre>