It appears that you may be missing a parentheses around (-2sqrt 3/2).
If that is the case, then using trig identities the answer should be 47.45 degrees.
Start by considering that the arccsc is 1/sin, sec is 1/cos, and cot is 1/tan.
+118609 Thanks anon.
I am having problems getting my head around this so I'd like another mathematician to take a look please.
acot(sec(acsc -2sqrt 3/2))
I also will interpret this as acot(sec(acsc( -2sqrt 3/2))) but you really did need brackets here.
$$acot(sec(acsc( \frac{-2\sqrt 3}{2})))\\\\
=acot(sec(acsc( -\sqrt 3))\\\\
=acot(sec(asin( \frac{1}{-\sqrt 3}))\\\\
NOTE \;\;asin( \frac{1}{-\sqrt 3})\quad $is an angle in the 4th quadrant$\\\\
$And sec of an angle in the 4th quad is positive$\\\\
=acot(\frac{\sqrt3}{\sqrt2})\\\\
=atan(\frac{\sqrt2}{\sqrt3})\\\\
=atan\sqrt{\frac{2}{3}}\\\\$$
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\sqrt{{\frac{{\mathtt{2}}}{{\mathtt{3}}}}}}\right)} = {\mathtt{39.231\: \!520\: \!483\: \!592^{\circ}}}$$
That is what I get but I'd really like someone else to look at it please.
Even if it is correct, is there an easier way of working it through?
ADDED
I ran this question through Wolfram|Alpha and got the same answer
+118609 Hi Alan,
You have not answered the same question as I did, and neither of us answered the question that was actually asked. I think that this is quite funny.
I'm going to try and answer the original question.
acot(sec(acsc -2sqrt 3/2))
I think technically this should be interpreted as;
$$\\acot(sec(\frac{acsc (-2) *\sqrt 3)}{2}))\\\\
=acot(sec(\frac{asin (\frac{1}{-2}) *\sqrt 3)}{2}))\\\\
=acot(sec(\frac{\frac{-\pi}{6} *\sqrt 3)}{2}))\\\\
=acot(sec(\frac{-\sqrt 3\pi}{12} ))\\\\$$
$${sec}{\left({\mathtt{\,-\,}}{\frac{\left({\frac{{\mathtt{180}}}{{\mathtt{\pi}}}}\right){\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{12}}}}\right)} = {\mathtt{1.112\: \!419\: \!829\: \!676\: \!128\: \!1}}$$
$${acot}{\left({sec}{\left({\mathtt{\,-\,}}{\frac{\left({\frac{{\mathtt{180}}}{{\mathtt{\pi}}}}\right){\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{12}}}}\right)}\right)} = {\mathtt{41.953\: \!677\: \!923\: \!85^{\circ}}}$$
Check with Wolfram|Alpha
