Following from this problem on time independence of Wronskian
Let $q:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. If $x_1(t)$ and $x_2(t)$ are the solutions to ODE: $\ddot{x}=q(t)x$ on $(a,b)$, we know that Wronskian determinate $$ \begin{vmatrix} x_1(t) & x_2(t)\\ \dot{x_1}(t)& \dot{x_2}(t) \end{vmatrix} $$
is time independent on $(a,b)$, since $\dot{W}=0$ in this case. If we assume that $$q(t)= \frac{3-2\sin(t) \cos(t)}{1+2\sin(t)\cos(t)}$$ and $a=0<b=\frac{\pi}{2}$, I have to verify that $$x_1(t) = \frac{1}{\sin(t)+\cos(t)}$$
is a solution to the ODE.
My initial try was to calculate the second derivative of $x_1$ and multiply it with the $q(t)$ but that got quite messy, is there another more elegant way to verify it?
Start with $x=v^{-1}$. Then $\dot x=-v^{-2}\dot v$ and $$ \ddot x=-v^{-2}\ddot v+2v^{-3}\dot v^2=\frac qv $$ On the other hand, $\ddot v=-v$ and so you need to test if $$ 1+2v^{-2}\dot v^2=q $$ Now apply $$ (\sin t\pm\cos t)^2=1\pm2\cos t\sin t $$