Question

If cosh x = 13 5 and x > 0, find the values of the other hyperbolic functions at x.

Answer 1

Recall that

$\cosh^2x-\sinh^2x=1\implies \cosh x=\pm\sqrt{1+\sinh^2x}$

You're given that $\cosh x=\dfrac{13}5>0$, so omit the negative root. Then

$\dfrac{13}5=\sqrt{1+\sinh^2x}\implies\sinh^2x=\dfrac{144}{25}\implies\sinh x=\pm\dfrac{12}5$

Because $x>0$, you have $\sinh x>0$ too, so omit the negative root again. Then $\sinh x=\dfrac{12}5$.

Now,

$\mbox{sech }x=\dfrac1{\cosh x}=\dfrac5{13}$
$\mbox{csch }x=\dfrac1{\sinh x}=\dfrac5{12}$
$\tanh x=\dfrac{\sinh x}{\cosh x}=\dfrac{\frac{12}5}{\frac{13}5}=\dfrac{12}{13}$
$\coth x=\dfrac1{\tanh x}=\dfrac{13}{12}$