Question 58223
<pre><font size = 5><b> 
    sinx           sinx
------------ + ------------
 (1 + cosx)     (1 - cosx)

The LCD is (1 + cosx)(1 - cosx)

Multiply the first fraction by 

(1 - cosx)         
---------- 
(1 - cosx)

which is OK to do since it equals 1:

    sinx(1 - cosx)           sinx
---------------------- + ------------
 (1 + cosx)(1 - cosx)     (1 - cosx)

Now multiply the second fraction by 

(1 + cosx)         
---------- 
(1 + cosx)

which is also OK to do since it equals 1:

    sinx(1 - cosx)           sinx(1 + cosx)
---------------------- + ----------------------
 (1 + cosx)(1 - cosx)     (1 - cosx)(1 + cosx)


Multiply everything out

    sinx - sinx·cosx        sinx + sinx·cosx
---------------------- + ----------------------
      1 - cos²x               1 - cos²x

Combine the numerators over the LCD:

 sinx - sinx·cosx + sinx + sinx·cosx
-------------------------------------
            1 - cos²x

the 2nd and 4th terms on top cancel out.
The 1st and 3rd terms on top combine to give 2sinx 

    2sinx 
-------------
  1 - cos²x

Use the identity sin²x + cos²x = 1, which
can be written sin²x = 1 - cos²x to replace
the bottom by sin²x

  2sinx 
---------
  sin²x

Divide top and bottom by sinx, and get

  2 
------
 sinx     

which can be written

 2cscx

Edwin</pre>