Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 8

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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