Question

Define $\sin hy$ and $\cos hy$ in terms of exponential functions and show that
$2y=\ln \left\{\frac{\cos hy+\sin hy}{\cos hy-\sin hy}\right\}$
By putting $\tan hy=\frac{1}{3}$, deduce that
$\tan h^{-1}\left(\frac{1}{3}\right)=\frac{1}{2}\ln 2$

Answer 1

We know that,

$sinhy=\frac{e^x-e^{-x}}{2}$

$coshy=\frac{e^x+e^{-x}}{2}$

$\Rightarrow ln\left(\frac{coshy+sinhy}{coshy-sinhy}\right)=ln\left(\frac{e^y}{e^{-y}}\right)$

$\Rightarrow ln\left(e^{2y}\right)=2y$

Let $tan{h}^{-1}\left(\frac{1}{3}\right)=y$

$\Rightarrow tanhy=\frac{1}{3}$

$\Rightarrow \frac{e^{2y}-1}{e^{2y}+1}=1/3$

$\Rightarrow e^{2y}=2$

$\Rightarrow 2y=ln2$

$\Rightarrow y=\frac{1}{2}ln2$