Problem source

The real variables $\theta$ and $u$ are related by the equation
$\tan\theta=\sinh u$ and $0\leqslant\theta<\frac{1}{2}\pi.$ Let
$v=\mathrm{sech}u.$ Prove that 
\begin{questionparts}
\item  $v=\cos\theta;$
\item $\dfrac{\mathrm{d}\theta}{\mathrm{d}u}=v;$ 
\item $\sin2\theta=-2\dfrac{\mathrm{d}v}{\mathrm{d}u}\quad$ and $\quad\cos2\theta=-\cosh u\dfrac{\mathrm{d}^{2}v}{\mathrm{d}u^{2}};$
\item ${\displaystyle \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2}=0.}$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
v &= \mathrm{sech} u \\
&=  \frac{1}{\mathrm{cosh } u} \\
&= \frac{1}{\sqrt{1+\mathrm{sinh}^2 u}} \tag{$u > 0$} \\
&= \frac{1}{\sqrt{1+\tan^2 \theta}} \\
&= \frac{1}{\sqrt{\mathrm{sec}^2 \theta}} \\
&= \cos \theta \tag{$0 < \theta < \tfrac{\pi}{2}$}
\end{align*}

\item \begin{align*}
 && \tan \theta &= \textrm{sinh} u \\
\underbrace{\Rightarrow}_{\frac{\d}{\d u}} && \sec^2 \theta \cdot \frac{\d \theta}{\d u} &= \cosh u \\
\Rightarrow && \frac{\d \theta}{\d u} &=\cosh u \cdot \cos^2 \theta \\
&&&= \frac{1}{v} \cdot v^2 \\
&&&=v
\end{align*}

\item \begin{align*}
\sin 2 \theta &= 2 \sin \theta \cos \theta \\
&= 2 \sin \theta \cdot \frac{\d \theta}{\d u} \\
&= -2 \frac{\d v}{\d \theta} \cdot \frac{\d \theta}{\d u} \tag{$\cos \theta = v$} \\
&= -2 \frac{\d v}{\d u}
\end{align*}


\begin{align*}
&& \sin 2 \theta &= -2 \frac{\d v}{\d u} \\
\underbrace{\Rightarrow}_{\frac{\d}{\d u}} && 2 \cos 2 \theta  \cdot \frac{\d \theta}{\d u} &=  -2 \frac{\d^2 v}{\d u^2} \\
\Rightarrow && \cos 2 \theta &= - \frac{\d^2 v}{\d u^2} \frac{1}{v} \\
&&&= -\frac{\d ^2v}{\d u^2} \cosh u
\end{align*}

\item \begin{align*}
&& \frac{\d u}{\d \theta} &= \frac{1}{v} \\
\Rightarrow && \frac{\d^2 u}{\d \theta^2} &= -\frac{1}{v^2} \frac{\d v}{\d \theta} \\
&&&= \frac{1}{v^2} \sin \theta
\\
&& \frac{\d v}{\d \theta} &= -\sin \theta \\
\Rightarrow && \frac{\d^2 v}{\d \theta^2} &= -\cos \theta \\
&&&= - v \\
\end{align*}
Therefore
\begin{align*}
 \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2} &= \frac{1}{v} \cdot \left (-v\right) + \left ( - \sin \theta \right ) \cdot \left (\frac{1}{v^2} \sin \theta \right) + \frac{1}{v^2} \\
&= -1 + \frac{1-\sin^2 \theta}{v^2} \\
&= -1 + \frac{\cos^2 \theta}{v^2} \\
&= -1 + 1 \\
&= 0
\end{align*}

\end{questionparts}