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

Expert Answers
lemjay eNotes educator| Certified Educator

`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.

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question