1. Prove this

Pl Prove this:

$\displaystyle \tan^{-1}(\cosh\theta - \sinh\theta) = \frac{1}{2}\sin^{-1}(sech\theta)$

2. Originally Posted by kjchauhan
Pl Prove this:

$\displaystyle \tan^{-1}(\cosh\theta - \sinh\theta) = \frac{1}{2}$

Rewrite the identity:
$\displaystyle 2\tan^{-1}(\cosh\theta - \sinh\theta) = \sin^{-1}(sech\theta)$

Take tan of both sides:

$\displaystyle \tan{(2\tan^{-1}(\cosh\theta - \sinh\theta))} = tan(\sin^{-1}(sech\theta))$

$\displaystyle \cosh\theta=\frac{1}{2}(e^x+e^{-x})$

$\displaystyle \sinh\theta=\frac{1}{2}(e^x-e^{-x})$

Hence:
$\displaystyle \cosh\theta - \sinh\theta=e^{-x}$

Our identity becomes:

$\displaystyle \tan{(2\tan^{-1}{e^{-x}})} = tan(\sin^{-1}(sech\theta)$

Now we can work with LHS only - by using the double angle formula:

$\displaystyle \tan{2A}=\frac{2tanA}{1-tan^2A}$
in our case $\displaystyle A=\tan^{-1}{e^{-x}}$

Hence LHS=$\displaystyle \frac{2e^{-x}}{1-e^{-2x}}=\frac{2}{e^x-e^{-x}}$

Now we simplify the RHS

$\displaystyle RHS=\tan(\sin^{-1}(sech\theta)$

$\displaystyle =\frac{\sin(\sin^{-1}(sech\theta)}{\cos(\sin^{-1}(sech\theta))}$

$\displaystyle =\frac{sech\theta}{\sqrt{1-\sin^2{\sin^{-1}{sech\theta}}}}$

$\displaystyle =\frac{sech\theta}{\sqrt{1-sech^2\theta}}$

$\displaystyle =\frac{sech\theta}{\sqrt{1-\frac{1}{\cosh^2\theta}}}$

$\displaystyle =\frac{sech\theta cosh\theta}{\sqrt{\cosh^2\theta-1}}$

Use the identity:

$\displaystyle \cosh^2\theta-\sinh^2\theta=1$

Hence:

$\displaystyle \cosh^2\theta-1=\sinh^2\theta$

So RHS becomes:

$\displaystyle =\frac{sech\theta cosh\theta}{\sqrt{\sinh^2\theta}}$

$\displaystyle =\frac{\frac{1}{cosh \theta}cosh \theta}{sinh\theta}$

$\displaystyle =\frac{1}{sinh \theta}$

$\displaystyle =\frac{2}{e^x-e^{-x}}$ which is the LHS