Need help with Mean Value Theorem Please! Have a question from a Calculus past paper and don't really know where to go with this question, any help would be appreciated. The question from the paper is:
Suppose that $0<a<b.$
Show that $\frac{\sinh b-\sinh a}{\cosh b- \cosh a}=\coth (c)$ for some $c\in(a,b)$. (Hint: Apply the mean value theorem with $f(x)=\sinh x-\lambda \cosh x$ for a suitable choice of $\lambda \in \mathbb{R}$.
Just looking for any input to help solve this question, Many Thanks.
 A: The mean value theorem explicitly states that there exists some $c \in (a,b)$ such that:
$$\begin{split}
f'(c) &= \frac{f(b) - f(a)}{b-a} \\
\cosh{c} - \lambda \sinh{c} &= \frac{f(b) - f(a)}{b-a}
\end{split}$$
Now, we need $\coth c$. That would be:
$$\coth c = \frac{1}{\sinh c}\frac{f(b) - f(a)}{b-a} + \lambda$$
If $f(b) = f(a)$, we could get rid of that extra $\sinh c$ term on the right hand side and we would simply get $\coth c = \lambda$. So let's do that:
$$\begin{split}f(a) &= f(b) \\
\sinh a - \lambda \cosh a &= \sinh b - \lambda \cosh b \\
\lambda \cosh b - \lambda \cosh a &= \sinh b - \sinh a \\
\lambda &= \frac{\sinh b - \sinh a}{\cosh b - \cosh a}
\end{split}$$
Thus, there exists some $c \in (a,b)$ such that:
$$\coth c = \frac{\sinh b - \sinh a}{\cosh b - \cosh a}$$
You could arrive at this more directly by using Cauchy's mean value theorem, which states that given functions $f,g$ continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists some $c \in (a,b)$ such that:
$$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$
Simply set $f(x) = \sinh x$ and $g(x) = \cosh h$ and we arrive at the result more directly.
A: Hint: Use the Cauchy Mean Value Theorem from http://en.wikipedia.org/wiki/Mean_value_theorem
