Question 1079426
<pre>
{{{tan(theta)=8=8/1=opposite/(adjacent)}}} so we draw a very steep
right triangle with the opposite = 8 and the adjacent = 1. 

{{{drawing(120,400,-1,2,-1,9,locate(.13,.4,theta),
locate(0,8,pi/2-theta),
triangle(0,0,1,0,1,8),locate(.4,0,1),locate(1.05,4.3,8) )}}}

The small sharp angle at the top is the complement of <font face="symbol">q</font>
{{{cot(pi/2-theta)=adjacent/(opposite)=8/1=8}}}

So {{{tan(theta) = 8 = cot(pi/2-theta) = 8}}}

"COtangent" is an abbreviation for "COmplement's tangent",
and cot(<font face="symbol">p/2-q</font>) IS <font face="symbol">q</font>'s COmplement's tangent. 
That's why both are equal.

The opposite side of <font face="symbol">q</font> is the adjacent side of <font face="symbol">p/2-q</font> and vice-versa.

Edwin</pre>