Question 1055186
<pre><b><font size = 1>
We draw an angle in the 4th quadrant like below.  Since the cotangent is the
adjacent over the opposite, or {{{x/y}}}, we make the given cotangent,-3, into a
fraction {{{-3/1}}}, but in quadrant IV, x goes right and y goes down, so the
numerator x is positive and the denominator y is negative, so we change the {{{-3/1}}} to {{{("" + 3)/(-1)}}} and make the numerator
+3 be the value of x and the denominator -1 be the value of y.

so we have:    
 
{{{drawing(270,100,-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(.5,.8,x=3),locate(1.05,-.7,y=-1),red(locate(.05,.75,theta),arc(0,0,.2,-.5,0,320)),
locate(.3,-.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=(3)^2+(-1)^2,r^2=9+1,r^2=10,r=sqrt(10))}}} {{{drawing(300,100,-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(.5,.8,x=3),locate(1.05,-.7,y=-1),red(locate(.05,.75,theta),arc(0,0,.2,-.5,0,320)),
locate(.17,-.9,r=sqrt(10)) )}}}


Now we need to know that 

the sine is the opposite over the hypotenuse or y/r, which is 
{{{(-1)/sqrt(10))}}} or {{{-sqrt(10)/10}}}. 

the cosine is the adjacent over the hypotenuse or x/r, which is 
{{{3/sqrt(10))}}} or {{{3sqrt(10)/10}}}. 

the tangent is the opposite over the adjacent or y/x, which is 
{{{(-1)/3}}} or {{{-1/3}}}. 

the secant is the hypotenuse over the adjacent or r/x, which is 
{{{sqrt(10)/3)}}}. 

the cosecant is the hypotenuse over the opposite or r/y, which is 
{{{sqrt(10)/(-1))}}} or {{{-sqrt(10)}}}. 

Edwin</pre>