SOLUTION: <pre> sinx sinx ----------- + ----------- (1 + cosx) (1 - cosx)</pre>

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Question 58223: 
    sinx           sinx
------------ + ------------
 (1 + cosx)     (1 - cosx)
Found 2 solutions by funmath, Edwin McCravy:
Answer by funmath(2933)   (Show Source): You can put this solution on YOUR website!
The LCD is (1+cos(x))(1-cos(x))



Pythagorean identity
Cancel the sin(x)
reciprocal identity
Happy Calculating!!!
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
 
    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
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