Question 1208769
<pre> 

Remember the Pythagorean identity:
  {{{1+cot^2(theta)=csc^2(theta)}}} or {{{csc^2(theta)-cot^2(theta)=1}}} 

Work with the left side only for a while:
 
{{{1/(csc(A) - cot(A)) - 1/sin(A)}}}{{{""=""}}}{{{ 1/sin(A) - 1/(csc(A) + cot(A))}}}

{{{(1/(csc(A) - cot(A)))*( (csc(A) + cot(A))/(csc(A) + cot(A))) - csc(A) }}}{{{""=""}}}{{{1/sin(A) - 1/(csc(A) + cot(A))}}}

{{{(csc^""(A) + cot^""(A))/(csc^2(A) - cot^2(A)) - csc^""(A) }}}{{{""=""}}}{{{ 1/sin(A) - 1/(csc(A) + cot(A))}}}

{{{(csc^""(A) + cot^""(A))/1^"" - csc^""(A) }}}{{{""=""}}}{{{ 1/sin(A) - 1/(csc(A) + cot(A))}}}

{{{csc^""(A) + cot^""(A) - csc^""(A) }}}{{{""=""}}}{{{ 1/sin(A) - 1/(csc(A) + cot(A))}}}

{{{cot^""(A) }}}{{{""=""}}}{{{ 1/sin(A) - 1/(csc(A) + cot(A))}}}

Now work with the right side only:

{{{cot^""(A) }}}{{{""=""}}}{{{ csc^""(A) - (1/(csc(A) + cot(A)))*( (csc(A) - cot(A))/(csc(A) - cot(A)))}}}

{{{cot^""(A) }}}{{{""=""}}}{{{ csc^""(A) - ((csc^""(A) - cot^""(A))/(csc^2(A) - cot^2(A)))}}}

{{{cot^""(A) }}}{{{""=""}}}{{{ csc^""(A) - ((csc^""(A) - cot^""(A))/(1^""))}}}

{{{cot^""(A) }}}{{{""=""}}}{{{ csc^""(A) - (csc^""(A) - cot^""(A)^"")}}}

{{{cot^""(A) }}}{{{""=""}}}{{{ csc^""(A) - csc^""(A) + cot^""(A)}}}

{{{cot^""(A) }}}{{{""=""}}}{{{ cot^""(A)}}}

Edwin</pre>