Differentiation with trigonometric and logarithmic functions

Question:
Here is a function of $x$.
\[ y=\alpha \cos x + 2 - \cos x \log\frac{1+\cos x}{1-\cos x} \]
$\alpha$ is a constant, and the range of $x$ is $0<x<\pi$.
Calculate $\frac{d^2 y}{dx^2}+\frac{\cos x}{\sin x}\frac{dy}{dx}$.

Answer:
Define each term.
\[ y_1 =  \alpha \cos x \\
    y_2 = 2 \\
    y_3 = \log\frac{1+\cos x}{1-\cos x}\]
Calculate each derivative.
\[ y_1' = -\alpha \sin x  \\
  y_2' = 0 \\  \]
The above derivatives are easy, but the third one is a little complicated:
\[  y_3' = (\log|1+\cos x| - \log|1-\cos x|)' \\
         = \frac{(1+\cos x)'}{1+\cos x} - \frac{(1-\cos x)'}{1-\cos x}   \\
         = \frac{-\sin x}{1+\cos x} - \frac{\sin x}{1-\cos x}   \\
        = \frac{-\sin x(1-\cos x) - \sin x(1+\cos x)}{1 - \cos^2x}  \\
        = \frac{-2\sin x}{\sin^2 x}   \\
        = \frac{-2}{\sin x} \]
Since $y=y_1+y_2-\cos x y_3$, the derivative becomes $y'=y_1'+y_2'-[(\cos x)' y_3+\cos x y_3']$. Namely,
\[ \frac{dy}{dx} = -\alpha\sin x + \sin x \log\frac{1+\cos x}{1-\cos x} + \frac{2\cos x}{\sin x}\]
The second derivative is calculated as follows:
\[ \frac{d^2y}{dx^2} = -\alpha\cos x + \cos x \log\frac{1+\cos x}{1-\cos x} + \sin x\frac{-2}{\sin x} + \frac{2((\cos x)'\sin x - \cos x(\sin x)')}{\sin^2x}  \\
           = -\alpha\cos x + \cos x \log\frac{1+\cos x}{1-\cos x} - 2 - \frac{2}{\sin^2x} \]
You can notice that the first three terms of above is equal to $-y$. Therefore,
\[ \frac{d^2y}{dx^2} = -y -\frac{2}{\sin^2 x} \]
The next term becomes
\[ \frac{\cos x}{\sin x}\frac{dy}{dx} = -\alpha\cos x + \cos x\log\frac{1+\cos x}{1-\cos x} + \frac{2\cos^2 x}{\sin^2 x} \]
This can be rewrite as
\[ \frac{\cos x}{\sin x}\frac{dy}{dx} = -y + 2 + \frac{2\cos^2 x}{\sin^2 x} \]
Thus, we have
\[ \frac{d^2 y}{dx^2}+\frac{\cos x}{\sin x}\frac{dy}{dx} = -2y\]

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