I'm trying to simplify the function
$$ \frac {76 \cosh(3x)}{1+\sinh(3x)^2} $$
I'm trying to get to the answer
$$ \frac{152e^{3x}}{(e^{3x})^2+1} $$
However I kept on getting the answer when trying to simplify the function $$ \frac{152(e^{3x}+e^{-3x})}{e^{6x}+e^{-6x}+2} $$
My graphing calculator graphed an identical graph when I typed the 2 function in, is my function incorrect or is there a way to get to that answer?
Any help would be appreciated, thanks in advance!
Via the identities $\cosh^2 x = 1 + \sinh^2x$ and $\cosh x = \frac{e^x+e^{-x}}{2} = \frac{e^{2x}+1}{2e^x}$, we easily get $$\frac{76\cosh 3x}{1+ \sinh^2 3x} = \frac{76}{\cosh 3x} = \frac{152e^{3x}}{e^{6x}+1}$$ which I think is what you wanted. On the other hand, $$e^{6x}+e^{-6x}+2 = (e^{3x}+e^{-3x})^2$$ so $$\frac{152(e^{3x}+e^{-3x})}{e^{6x}+e^{-6x}+2} = \frac{152}{e^{3x}+e^{-3x}} = \frac{152e^{3x}}{e^{6x}+1}$$ The two answers are indeed the same.
Let us drop these useless factors $76$ and $3$.
$$\frac{\cosh x}{1+\sinh^2x}=\frac{2e^x}{(e^x)^2+1}$$
because
$$\frac{\cosh x}{\cosh^2x}=\frac{2}{e^x+e^{-x}}.$$