Question

if tanh(x) =5/13 , find the value of the hyperbolic function of cosh(x) ...?

Answer 1

$\tanh(x)=\frac{\sinh(x)}{\cosh(x)}\rightarrow \sinh(x)=\cosh(x)\tanh(x)$

Now,

$\cosh^2(x)-\sinh^2(x)=1$

so substituting for sinh(x):

$\cosh^2(x)-\cosh^2(x)\tanh^2(x)=1\rightarrow \cosh^2(x)(1-\tanh^2(x))=1$

which means

$\cosh(x)=\pm\sqrt{\frac{1}{1-\tanh^2(x)}}=\pm\sqrt{\frac{1}{1-(\frac{5}{13})^2}}=\pm\sqrt{\frac{1}{1-\frac{25}{169}}}=\pm\sqrt{\frac{1}{\frac{144}{169}}}= \pm\frac{13}{12}$