[1/(secx+1)]-[1/(secx-1)] =-2cot2x

Question

Could someone please show me how to do a step by step proof of this problem with each method listed out please?

Answer 1

Do you mean -2cot²x on the right side?  I think you must.

 

Look at the left hand side.  It is the different between two fractions: 1/(secx+1) minus 1/(secx-1).  Let's turn it into a single fraction with a common denominator.  The common denominator will be

(secx - 1)(secx + 1).

 

So multiply the first term by (secx - 1)(secx - 1) and multiply the second term by (secx + 1)(secx + 1), then combine the terms:

 

    1           -      1          =   (secx - 1) - (secx + 1)   =      -2        

secx + 1       secx - 1          (secx - 1) (secx + 1)         sec²x - 1

 

 

But sec²x = 1 + tan²x, so the above fraction is equal to

 

    -2                    =  -2       = -2 cot²x, which is the same as the right side.

(1 + tan²x) - 1       tan²x

 

 

The only trick is that you might not already know that sec2x = 1 + tan²x.

 

To see that, remember that sin²x + cos²x = 1

 

Divide both sides by cos²x:
sin²x/cos²x + cos²x/cos²x = 1/cos²x

 

That's the same as:
(sin x/cos x)² + 1 = (1/cos x)²

 

But sinx / cosx = tan x, and 1/cosx = sec x

 

So that's the same as

tan²x + 1 = sec²x

 

So you can use tan²x + 1 in place of sec²x above.

 

Answer 2

This time, it does.

 

       1                  1

-------------  -  ------------   =   -2cot²x

(sec x + 1)      (sec x - 1)

 

 

(sec x - 1) - (sec x + 1)                 -2

------------------------------         =  --------   = -2cot²x     

         sec²x - 1                             tan²x

 

 

 

There you have it.  Good luck to you.

 

(Note: Don't mess with signs.)