Linear independence of solutions (sorry, there was a typo in my original post - it has been fixed now)
$$y''+\left(\frac{\sin x}{1+\cos x}\right)y'+\left(\frac{1}{1+\cos x}\right)y=0$$
I can see that both $A\sin x$ and $B\cos x+B$ are solutions. The Wronskian is:
$$W(x)=\text{det}\begin{pmatrix} 1+\cos x\quad \sin x\\ -\sin x\quad \cos x\end{pmatrix}=1+\cos x$$
This is $0$ at $\pi$; I am asked to comment on the solution set.
Thoughts: Despite $W$ being $0$ somewhere it is certainly not identically $0$. I put this down to $\pi$ being a singularity point of the equation, so it technically doesn't lie in the domain of $W$. Is this a good reason?
We know $\sin $ and $\cos $ are linearly independent so $A\sin x+B\cos x+B$ must in fact be the most general solution. Is this true/correct? Wolfram seems to disagree.
 A: The equation
$$
   (1+\cos x)y''+\sin x\; y' + y = 0.
$$
has one solution $y=1+\cos x$ because $y'=-\sin x$, $y''=-\cos x$ gives
$$
         -(\cos x +\cos^{2}x) -\sin^{2}x + (1+\cos x) = 0.
$$
The second solution can be found by variation of parameters. Let $y_{1}=1+\cos x$ and $y_{2}=a(x)y_{1}$ for a function $a(x)$ to be found:
$$
            (1+\cos x)(ay_{1}''+2a'y_{1}'+a''y_{1})+\sin x(ay_{1}'+a'y_{1})+ay_{1}=0.
$$
Using $(1+\cos x)(ay_{1}'')+\sin x(ay_{1})+ay_{1}=0$ gives
$$
    (1+\cos x)(2a'y_{1}'+a''y_{1})+\sin x(a'y_{1})=0, \\
   a''y_{1}(1+\cos x)+a'\{2y_{1}'(1+\cos x)+y_{1}\sin x \}=0,\\
       \frac{a''}{a'}+\left(\frac{2y_{1}'}{y_{1}}+\frac{\sin x}{1+\cos x}\right)=0,\\
             \ln |a'| + \ln y_{1}^{2}-\ln|1+\cos x|= C,\\
              a' = D\frac{1+\cos x}{y_{1}^{2}}=D\frac{1}{1+\cos x}.
$$
So a second solution is $y_{2}=ay_{1}$, where $a=1+\cos x$:
$$
                   y_{2}=\int_{0}^{x}\frac{1}{1+\cos u}\,du\cdot (1+\cos x).
$$
The general solution is $Ay_{1}+By_{2}$, where $A$, $B$ are constants.
The Wronskian is
$$
     W(y_{1},y_{2})=W(y_{1},ay_{1})=y_1(a'y_1+ay_1')-y_1'(ay_1)=a'y_1^{2}=1+\cos x.
$$
As you stated, the Wronskian is not technically defined at $x=\pi$ because $y_{2}$ is not defined there. This is a limiting value of the Wronskian, and $0$ is okay as a limiting value.
