Question 832233
<pre>
If {{{tan(theta) = 2}}} and {{{sin(theta) < 0}}}, what would the exact value of {{{csc(theta)}}} be?


The tangent is positive (since it's given as +2) and the sine is negative,
(since we are given {{{sin(theta)<0}}}. The tangent is positive in the
1st and 3rd quadrants, and the sine is negative in the 3rd and 4th, so
{{{theta}}} must be in the 3rd quadrant, or Q3.  So we draw an angle in
the 3rd quadrant like below.  Since the tangent is the opposite over the
adjacent, or {{{y/x}}}, we make the given tangent, 2, into a fraction {{{2/1}}}, but 
x goes left and y goes down, so they both have to be negative, so we
change the {{{2/1}}} to {{{(-2)/(-1)}}} and make the numerator -2 be the value of y
and the denominator -1 be the value of x, so we have:    
 
{{{drawing(300,300,-2,2,-2,2,line(-30,0,30,0),line(0,-30,0,30),
green(line(0,0,-1,-2),line(-1,0,-1,-2)),locate(-.8,.3,x=-1),locate(-1.6,-.8,y=-2),red(locate(.05,.35,theta),arc(0,0,.4,-.4,0,240)),
locate(-.45,-.9,"r=?") )}}}{{{matrix(16,1,

But, we, "don't", know, r=hypotenuse, so, we, use, the, Pythagorean, theorem,
r^2=x^2+y^2,r^2=(-1)^2+(2)^2,r^2=1+4,r^2=5,r=sqrt(5))}}}{{{drawing(300,300,-2,2,-2,2,line(-30,0,30,0),line(0,-30,0,30),
green(line(0,0,-1,-2),line(-1,0,-1,-2)),locate(-.8,.3,x=-1),locate(-1.6,-.8,y=-2),red(locate(.05,.35,theta),arc(0,0,.4,-.4,0,240)),

locate(-.5,-1,r=sqrt(5)) )}}}

Now we just need to know that the cosecant is the hypotenuse over the
opposite or r/y, which is {{{(sqrt(5))/(-2)}}} or {{{-sqrt(5)/2}}}. 

{{{csc(theta)}}}{{{""=""}}}{{{(-1)/sqrt(5)}}} = {{{(-1sqrt(5))/(sqrt(5)sqrt(5))=-sqrt(5)/5}}}
 
Edwin</pre>