[1/(secx+1)]+[1/(secx-1)] = -2cot^2x

Question

Could someone please show me a step by step proof of this problem with all of the step details listed?

Answer 1

The given equation is not an identity.  

 

For example, when x = 30º, the left hand side of the equation is positive, but the right hand side of the equation is negative.  

Answer 2

Hi Blade,

 

Are you sure about [1/(secx+1)]+[1/(secx-1)] = -2cot²x?

 

      1              1         

----------- + --------- = -2cot²x

sec x + 1     sec x -1

 

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

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

           sec²x - 1

 

  2sec x                             2sec x

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

sec2x - 1                            tan²x

 

2sec x cot²x = -2cot²x   

(From this stand point, why your problem canceled sec x and why is your answer negative?)

 

2sec x cot²x = -2cot²x           can be written as:

 

   2     cos²x      2 cos x      2 cos x      1

------  -------  =  ---------   = ----------   --------   = 2 cot x csc x

cos x  sin²x         sin²x         sin x     sin x

 

Either (2sec x cot²x = -2cot²x) or  (2 cot x csc x = -2cot²x), no negative sign can be found.  Not only that, it doesn't match or it can't be verified.

 

I'm agreeing with Matt M on this one.  Check your problem again.

 

 

 

(For future note, d/dx  cot x = -csc²x      &     d/dx   csc x =  -csc x cot x)  d/dx = derivative

 

Answer 3

[1 / (sec + 1)] + [1 / (sec - 1)] = -2cot²                             Common denominator

(sec - 1 + sec + 1) / (sec + 1)(sec - 1) = -2cot²

(sec - 1 + sec + 1) / (sec² - 1) = -2cot²                              Simplify

2sec / (sec² - 1) = -2cot²                                           sec² - 1 = tan²

2sec / tan² = -2cot²                                                   1 / tan² = cot²

2sec (cot²) = -2cot²                                                                   

 

Three possibilities:

1) I have made an error that I cannot find;

2) The identity is not true;or,

3) Some of the given values are incorrect.