Question 854305
<pre>
You were to sketch a triangle, not a circle.

cot(&#952;)=1

The cotangent is the adjacent over the opposite.

Write the 1 as {{{1/1}}}

{{{cot(theta)=1/1}}}

Then when we draw the right triangle, we will make the adjacent side
to {{{theta}}} equal to the numerator of {{{1/1}}}, which is 1 and the opposite
side of {{{theta}}} equal to the denominator of {{{1/1}}} which is 1 as well, so the
right triangle is:

{{{drawing(300,300,-1.5,1.5,-.5,2.5, triangle(-1,0,1,0,1,2),locate(-.6,.26,theta),red(arc(-1,0,1.7,-1.7,0,45)),
locate(0,-.1,1),locate(1.1,1,1) )}}}

Now we calculate the hypotenuse by the Pythagorean theorem:

{{{hypotenuse^2=adjacent^2+opposite^2}}}

{{{hypotenuse^2=1^2+1^2}}}

{{{hypotenuse^2=1+1}}}

{{{hypotenuse^2=2}}}

{{{sqrt(hypotenuse^2)=sqrt(2)}}}

{{{hypotenuse = sqrt(2)}}}

{{{drawing(300,300,-1.5,1.5,-.5,2.5, triangle(-1,0,1,0,1,2),locate(-.6,.26,theta),red(arc(-1,0,1.7,-1.7,0,45)),locate(-.1,1.2,sqrt(2)),
locate(0,-.1,1),locate(1.1,1,1) )}}}

Now the adjacent side = 1, the opposite side = 1, 
and the hypotenuse = {{{sqrt(2)}}}

{{{sin(theta)=opposite/hypotenuse=1/sqrt(2)=sqrt(2)/2}}}
{{{cos(theta)=adjacent/hypotenuse=1/sqrt(2)=sqrt(2)/2}}}
{{{tan(theta)=opposite/adjacent=1/1=1}}}
{{{sec(theta)=hypotenuse/adjacent=sqrt(2)/1=sqrt(2)}}}
{{{csc(theta)=hypotenuse/opposite=sqrt(2)/1=sqrt(2)/2}}}

And you were already given:

{{{cot(theta)=adjacent/opposite=1/1=1}}}

Edwin</pre>