Question 1191899
<pre>
{{{cot(tan(x)^"")}}}{{{""=""}}}{{{tan(cot(x)^"")}}}

{{{cos(tan(x)^"")/sin(tan(x)^"")}}}{{{""=""}}}{{{sin(cot(x)^"")/cos(cot(x)^"")}}}

Cross-multiply:

{{{cos(tan(x)^"")cos(cot(x)^"")}}}{{{""=""}}}{{{sin(cot(x)^"")sin(tan(x)^"")}}}

Get 0 on the right side:

{{{cos(tan(x)^"")cos(cot(x)^"")-sin(cot(x)^"")sin(tan(x)^"")}}}{{{""=""}}}{{{0}}}

Use the identity {{{cos(alpha+beta)}}}{{{""=""}}}{{{cos(alpha)cos(beta)-sin(alpha)sin(beta)}}}

{{{cos(tan(x)+cot(x)^"")}}}{{{""=""}}}{{{0}}}

So tan(x)+cot(x) must equal to any odd integer multiple of {{{pi/2}}}

The general odd integer is 2n+1 for any integer n

{{{tan(x)+cot(x)}}}{{{""=""}}}{{{(2n+1)(pi/2)}}}

{{{sin(x)/cos(x)+cos(x)/sin(x)}}}{{{""=""}}}{{{(2n+1)(pi/2)}}}

{{{(sin^2(x)+cos^2(x))/(cos(x)sin(x))}}}{{{""=""}}}{{{(2n+1)(pi/2)}}}

{{{1/(cos(x)sin(x))}}}{{{""=""}}}{{{(2n+1)(pi/2)}}}

Take reciprocals of both sides:

{{{cos(x)sin(x)}}}{{{""=""}}}{{{1/((2n+1)(pi/2))}}}

Multiply numerator and denominator on the right side by 2

{{{cos(x)sin(x)}}}{{{""=""}}}{{{2/((2n+1)pi)}}}

Multiply both sides by 2 and reverse the factors on the left side
to make the left side look like the right side of the 
identity {{{sin(2x)}}}{{{""=""}}}{{{2sin(x)cos(x)}}}

{{{2sin(x)cos(x)}}}{{{""=""}}}{{{4/((2n+1)pi)}}}

{{{sin(2x)}}}{{{""=""}}}{{{4/((2n+1)pi)}}}, where n is any integer.

Edwin</pre>