I am given the question:
Using the determinant, show that the 1+1d Lorentz transformation matrix $\Lambda$ can be written in terms of hyperbolic trig functions, $$ \Lambda = \begin{pmatrix} \cosh u & -\sinh u \\ -\sinh u & \cosh u \end{pmatrix} . $$
This seems like a pretty paltry hint. Sure, the determinant of the usual 1+1d Lorentz matrix is $1 - \beta^2$, but how does that even remotely help?
Suppose that
$$ \Lambda = \begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \end{pmatrix} $$
where $\beta=\frac{v}{c}$ and $\gamma = (1-\beta^2)^{-\frac{1}{2}}$. Note that $\beta \in (-1, 1)$, so we can define $u$ as the unique real number such that $\beta = \tanh u$. Now, $\det \Lambda = \gamma^2 - \beta^2\gamma^2 = \gamma^2(1 - \beta^2) = 1$ so
$$ \gamma^2 = \frac{1}{1 - \tanh^2 u} = \cosh^2 u, $$
but $\cosh$ and $\gamma$ are positive so $\gamma = \cosh u$. Therefore,
$$ \Lambda = \begin{pmatrix} \cosh u & -\sinh u \\ -\sinh u & \cosh u \end{pmatrix} $$
as expected.
