Monday, November 5, 2012

`coth^2(x) - csc h^2(x) = 1` Verify the identity.

`coth^2(x) - csc h^2(x) =1`


Take note that hyperbolic cotangent and hyperbolic cosecant are defined as


  • `coth (x) = (e^x+e^(-x))/(e^x-e^(-x))`

  • `csc h^2(x) =2/(e^x - e^(-x))`

Plugging them, the left side of the equation becomes


`((e^x+e^(-x))/(e^x-e^(-x)))^2 -(2/(e^x - e^(-x)) )^2=1`


`(e^x+e^(-x))^2/(e^x-e^(-x))^2 -2^2/(e^x - e^(-x))^2=1`


`(e^x+e^(-x))^2/(e^x-e^(-x))^2 -4/(e^x - e^(-x))^2=1`


`((e^x+e^(-x))^2-4)/(e^x - e^(-x))^2=1`


Then, simplify the numerator.


`((e^x + e^(-x))(e^x + e^(-x)) - 4)/(e^x- e^(-x))^2=1`


`(e^(2x)+1+1+e^(-2x) - 4)/(e^x- e^(-x))^2=1`


`(e^(2x)+2+e^(-2x) - 4)/(e^x- e^(-x))^2=1`


`(e^(2x) - 2 +e^(-2x)) /(e^x- e^(-x))^2=1`


Factoring the numerator, it becomes


`((e^x - e^(-x))(e^x-e^(-x)))/(e^x- e^(-x))^2=1`


`(e^x - e^(-x))^2/(e^x- e^(-x))^2=1`


Cancelling common factor, the right side simplifies to


`1=1`


This verifies that the given equation is an identity.



Therefore,  `coth^2(x) - csc h^2(x)=1`  is an identity.

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