# Hyperbolic Trig Functions - Identities

I don't understand how the 3rd step (the 4 divisions) happens? Can someone explain how they arrived at that.

• By dividing both numerator and denominator by cosh $x$ cosh $y$ – Extremal Mar 3 '15 at 2:43

$$1=\frac{\frac1{\cosh x \cosh y}}{\frac1{\cosh x \cosh y}}$$
$$\frac{\sinh x\cosh y+\cosh x\sinh y}{\cosh x\cosh y+\sinh x\sinh y}=\left(\frac{\sinh x\cosh y+\cosh x\sinh y}{\cosh x\cosh y+\sinh x\sinh y}\right)\left(1\right)=\left(\frac{\sinh x\cosh y+\cosh x\sinh y}{\cosh x\cosh y+\sinh x\sinh y}\right)\left(\frac{\frac1{\cosh x \cosh y}}{\frac1{\cosh x \cosh y}}\right)=\frac{\frac{\sinh x\cosh y}{\cosh x \cosh y}+\frac{\cosh x\sinh y}{\cosh x \cosh y}}{\frac{\cosh x\cosh y}{\cosh x \cosh y}+\frac{\sinh x\sinh y}{\cosh x \cosh y}}$$