The other tutor mistakenly took your word "exact" 
for "approximate". 

The convention for inverse tangent is:

0 < tan-1(positive_number) < 

tan-1(0) = 0

 < tan-1(negative_number) < 0

Therefore in your problem:

sin[tan-1(-√3)]

ee first find the tan-1(-√3).

That asks the question:

"What angle in the fourth quadrant taken in the interval (,0)
has a tangent of -√3?"

That brings to mind the special angle 60° or , whose tangent is +√3.
So we think of an angle in quadrant IV that has 60° for its referent angle,
since angles in Q4 have negative tangents.

[We may think of 300° or , even though by convention it's actually
 measured clockwise as .

Anyway,

sin[tan-1(-√3)] = sin() = .
It's negative since the sine is negative in QIV.

Answer:  

Edwin