Question

How can you prove that `1 - tanh^2x = sech^2x`?

Answer 1

Let's rewrite in terms of `e^x`, using the identities `sechx = 2/(e^x + e^-x)` and `tanhx = (e^x - e^-x)/(e^x + e^-x)`.

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

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

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

`1- (e^(2x) - e^0 - e^0 + e^(-2x))/(e^(2x) + e^0 + e^0 + e^(-2x)) = 4/(e^(2x) + e^0 + e^0 + e^(-2x))`

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

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

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

This has been proved!

Hopefully this helps!