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

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Answer 1 · 2017-02-07T19:05:19

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!

How can you prove that 1 - tanh^2x = sech^2x1 - tanh^2x = sech^2x?