Help with hyperbolic trig I have been working on a problem in my physics homework and I am stuck on a particular line of many trying to show that the following hyperbolic trig expressions are equivalent.  Can someone please help?  I want to show:
$$ -\frac{1}{2}(x^2+y^2)\coth(\gamma)+\frac{xy}{\sinh(\gamma)}=-\frac{1}{4}\left[(x+y)^2\tanh\left(\frac{\gamma}{2}\right)+(x-y)^2\coth\left(\frac{\gamma}{2}\right)\right] $$
where I am told this identity has been used:
$$ \tanh\left(\frac{\gamma}{2}\right)=\frac{\cosh(\gamma)-1}{\sinh(\gamma)}=\frac{\sinh(\gamma)}{1+\cosh(\gamma)} $$
I am just going in circles and getting lost in the steps. Please help!  Thank you!
 A: After writing the problem on the screen, things seemed to become clear and I found a missing factor of 4 in a defined constant. Now everything seems okay. I start with the expression in brackets on the right-hand side and rewrite using the given indentities.
$$ -\left[(x+y)^2\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+(x-y)^2\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$
$$ =-\left[(x^2+2xy+y^2)\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+(x^2-2xy+y^2)\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$
$$ =-\left[(x^2+y^2)\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+\frac{2xy\sinh(\gamma)}{1+\cosh(\gamma)}+(x^2+y^2)\frac{\sinh(\gamma)}{\cosh(\gamma)-1}-2xy\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$
$$ =-\left[(x^2+y^2)\left(\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right)+2xy\frac{\sinh(\gamma)}{\cosh+1}-2xy\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$
Take last two terms on the right.
$$ 2xy\frac{\sinh(\gamma)}{\cosh(\gamma)+1}-2xy\frac{\sinh(\gamma)}{\cosh(\gamma)-1} $$
$$ =\frac{2xy\sinh(\gamma)(\cosh(\gamma)-1)-2xy\sinh(\gamma)(\cosh(\gamma)+1)}{(\cosh(\gamma)+1)(\cosh(\gamma)-1)} $$
$$ =4xy\frac{\sinh(\gamma)}{\cosh^2(\gamma)-1}=-\frac{4xy}{\sinh(\gamma)} $$
Now look at first two terms on the left of the expression.
$$ -(x^2+y^2)\left[\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$
$$ =-(x^2+y^2)\left[\frac{\sinh(\gamma)\cosh(\gamma)-\sinh(\gamma)+\sinh(\gamma)+\sinh(\gamma)\cosh(\gamma)}{\sinh^2(\gamma)}\right] $$
$$ =-2(x^2+y^2)\coth(\gamma) $$
Then plugging it all, we have
$$ -\frac{1}{2}(x^2+y^2)\coth(\gamma)+\frac{xy}{\sinh(\gamma)}=-\frac{1}{4}\left[2\coth(\gamma)-\frac{4xy}{\sinh(\gamma)}\right] $$
and we arrive at the desired result.  Thank everyone for their useful comments!
