Question 1003700
<pre>
{{{2cot(A) = tan(B) - cot(B)}}}

We work with the left side.

given isosceles triangle ABC, with B = C we have:

A+B+C = 180°
A+B+B = 180°
A+2B = 180°
A = 180° - 2B
cot(A) = cot(180°-2B) = -cot(2B)

So the left side

{{{2cot(A)}}}  becomes

{{{-2cot(2B)}}}

{{{-2(cos(2B)/sin(2B))}}}

{{{-2((cos^2(B)-sin^2(B))/(2sin(B)cos(B)))}}}

{{{-cross(2)((cos^2(B)-sin^2(B))/(cross(2)sin(B)cos(B)))}}}

{{{-((cos^2(B)-sin^2(B))/(sin(B)cos(B)))}}}

{{{(-cos^2(B)+sin^2(B))/(sin(B)cos(B))}}}

{{{(-cos^2(B))/(sin(B)cos(B))}}}{{{""+""}}}{{{(sin^2(B))/(sin(B)cos(B))}}}

{{{(-cos^cross(2)(B))/(sin(B)cross(cos(B)))}}}{{{""+""}}}{{{(sin^cross(2)(B))/(cross(sin(B))cos(B))}}}

{{{-cos(B)/sin(B)+sin(B)/cos(B)}}}

{{{-cot(B)+tan(B)}}}

{{{tan(B)-cot(B)}}}

Edwin</pre>