Question 891554
<pre>
{{{csc(theta)}}}{{{""=""}}}{{{2&1/2}}}

Change {{{2&1/2}}} to an improper fraction:

{{{csc(theta)}}}{{{""=""}}}{{{5/2}}}

Since {{{cosecant=hypotenuse/opposite}}} we draw a right
triangle with an angle {{{theta}}} with a hypotenuse of 5, which is 
the same as the numerator of the fraction {{{5/2}}}, and with 
a side opposite angle {{{theta}}} of 2, which is the denominator
of the fraction {{{5/2}}}.

{{{drawing(400,2400/11,-.5,5,-.5,2.5,

triangle(0,0,sqrt(21),0,sqrt(21),2), locate(.6,.25,theta),locate(4.61,1,2),
locate(2.2,1.2,5) )}}}

[You don't have to be accurate with your drawing, just any sketch of
a right triangle will do.]

Next we use the Pythagorean theorem to calculate the adjacent side
to angle {{{theta}}}:

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

{{{adjacent^2+2^2=5^2}}}

{{{adjacent^2+4=25}}}

{{{adjacent^2=21}}}

{{{adjacent=sqrt(21)}}}

So we label the adjacent side as {{{sqrt(21)}}}

{{{drawing(400,2400/11,-.5,5,-.5,2.5,
locate(2.2,-.1,sqrt(21)),
triangle(0,0,sqrt(21),0,sqrt(21),2), locate(.6,.25,theta),locate(4.61,1,2),
locate(2.2,1.2,5) )}}}

{{{sin(theta)=opposite/hypotenuse=2/5}}}

{{{cos(theta)=adjacent/hypotenuse=sqrt(21)/5}}}

{{{tan(theta)=opposite/adjacent=2/sqrt(21)=expr(2/sqrt(21))*red(sqrt(21)/sqrt(21))=2sqrt(21)/21}}}

{{{cot(theta)=adjacent/opposite=sqrt(21)/2}}}

{{{sec(theta)=hypotenuse/adjacent=5/sqrt(21)=expr(5/sqrt(21))*red(sqrt(21)/sqrt(21))=5sqrt(21)/21}}}

Edwin</pre>