Simplify $Y_n(y) = \cosh{\pi ny} - \coth{\pi n}\sinh{\pi n y}$ into the form $Y_n(y) = A(\pi n B)$ for functions $A$ and $B$, such that $B > 0$ for $y > 1$
So, I've managed to simplify the expression into a single exponential, however, it does not satisfy the condition I've been given for B. An help on how to do this would be very much appreciated.
My Solution: Let $\pi n = w$, then $$ \begin{align} Y_n(y) &= \frac{e^{wy}+e^{-wy}}{2} - \frac{e^w+e^{-w}}{e^{w} - e^{-w}} \cdot \frac{e^{wy} - e^{-wy}}{2}\\ &= \frac{(e^{wy} + e^{-wy})(e^w - e^{-w}) - (e^w - e^{-w})(e^{wy} - e^{-wy})}{2(e^w - e^{-w})}\\ &= \frac{2e^{w(1-y)} - 2e^{-w(1+y)}}{2(e^w - e^{-w})}\\ &= \frac{e^{-wy-w}(e^{2w} - 1)}{e^{-w}(e^{2w} - 1)}\\ &= e^{-wy} \end{align} $$
Which gives $A(y) = \exp(\pi n B)$ with $B(y) = -y$. But, of course this gives $B < 0$ for $y > 1$
$e^{-wy}=A(wy)$ for $A(t):=e^{-t}.$ But this is not the correct answer, because there is mistake in you calculation: $(e^{w} - e^{-w})$ in the second numerator became $(e^{w} + e^{-w})$ when you transformed the difference into a single fraction.
$$Y_n(y)=\frac{\sinh(w(1-y))}{\sinh w}=A(wB)$$ for $$A(t):=-\frac{\sinh(wt)}{\sinh w},\quad B(y):=y-1.$$