Question 550824
True.
For all x, {{{cot(x)=cos(x)/sin(x)}}} and {{{csc(x)=1/sin(x)}}}
So, for all x, 
cot^2x-csc^2x={{{(cot(x))^2-(csc(x))^2}}} = {{{(cos(x)/sin(x))^2-(1/sin(x))^2}}} =  {{{(cos(x))^2/(sin(x))^2-1/(sin(x))^2}}} = {{{((cos(x))^2-1)/(sin(x))^2}}} = {{{-(1-(cos(x))^2)/(sin(x))^2}}} = {{{-(sin(x))^2/(sin(x))^2}}} = -1