A familiar identity is  tan = , and
since the cotangent is the reciprocal of the tangent, we have
cot = ,  so we will try to make
the left side into that expression. So we will multiply the left side
by , so hopefully it will have the desired denominator in the end. 

 
1 + sin(x) + cos(x)
——————————————————— = cot()
1 + sin(x) - cos(x)  


[1 + sin(x) + cos(x)]   [1 - cos(x)] 
————————————————————— · ————————————
[1 + sin(x) - cos(x)]   [1 - cos(x)]

1 - cos(x) + sin(x) - sin(x)cos(x) + cos(x) - cos²(x) 
—————————————————————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

1 - cos(x) + sin(x) - sin(x)cos(x) + cos(x) - cos²(x) 
—————————————————————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

      1 + sin(x) - sin(x)cos(x) - cos²(x) 
      ———————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

Rearrange the terms

      1 - cos²(x) + sin(x) - sin(x)cos(x)  
     —————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

     [1 - cos²(x)] + sin(x) - sin(x)cos(x)  
     —————————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]
  
Use the identity sin²(x) = 1 - cos²(x)

        sin²(x) + sin(x) - sin(x)cos(x)  
       —————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

Factor out sin(x) on the top:

         sin(x)[sin(x) + 1 - cos(x)]  
       —————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]

Then we can cancel:

         sin(x)[sin(x) + 1 - cos(x)]  
       —————————————————————————————————
       [1 + sin(x) - cos(x)][1 - cos(x)]
 
                  sin(x)  
                ——————————
                1 - cos(x)

which is the identity we showed above for cot()

                cot

Edwin