<s>[INST] Having trouble understanding Re-derivation of Young’s Equation I was going over a section 2.1 of this article (regarding Young's Equation and for some reason I wasn't able to derive E10 from E9 (see image below). I know it is more about the trigonometry of the problem, but when I tried to derive it myself it became cumbersome real fast and I don't get to E10. I would appreciate if someone could help me with that.

 [/INST] The first parenthesis in E9 is simplified into
$$
\begin{aligned}
&\sin^2\theta + \cos\theta \frac{-(1 - \cos\theta)(2 + \cos\theta)}{1 + \cos\theta} \\
= &(1 - \cos^2\theta) - \cos\theta \frac{2 - \cos\theta - \cos^2\theta}{1 + \cos\theta} \\
= &1 - \cos\theta \frac{\cos\theta(1 + \cos\theta) + (2 - \cos\theta - \cos^2\theta)}{1 + \cos\theta} \\
= &1 - \cos\theta \frac{2}{1 + \cos\theta} \\
= &\frac{1 - \cos\theta}{1 + \cos\theta}
\end{aligned}
$$
The second parenthesis in E9 is simplified into
$$
\begin{aligned}
&2(1 - \cos\theta) - \frac{(1 - \cos\theta)(2 + \cos\theta)}{1 + \cos\theta} \\
= &\frac{2 (1-\cos\theta)(1+\cos\theta) - (1-\cos\theta)(2+\cos\theta)}{1+\cos\theta} \\
= &\frac{2 - 2\cos^2\theta - 2 + \cos\theta + \cos^2\theta}{1+\cos\theta} \\
= &\frac{-\cos^2\theta + \cos\theta}{1 + \cos\theta} \\
= &\cos\theta \frac{1 - \cos\theta}{1 + \cos\theta}
\end{aligned}
$$
Now use these simplifications in the E9
$$(\gamma_{sl} - \gamma_{so}) \frac{1 - \cos\theta}{1 + \cos\theta} + \gamma \cos\theta \frac{1 - \cos\theta}{1 + \cos\theta} = 0$$
$$\frac{1 - \cos\theta}{1 + \cos\theta} \Bigl( (\gamma_{sl} - \gamma_{so}) + \gamma \cos\theta \Bigr) = 0$$
I do not know specifics about the above equation, but from purely mathematical point of view it has a singularity at $\theta = \pi$, and for $\theta = 0$ you cannot assume the term in parenthesis to be equal to zero.
If we neglect these two special cases for $\theta$ the Eq. E9 is simplified into
$$\boxed{\gamma_{sl} + \gamma \cos\theta = \gamma_{so}} \tag {E10}$$
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