Simplifying trigonometric identities

$\begin{aligned} \frac{1+\sin(x)-\cos(x)}{1+\sin(x)+\cos(x)} &= \frac{1-\cos(x)+\sin(x)}{1+\cos(x)+\sin(x)} \\ &= \frac{\frac{1-\cos(x)}{\sin(x)} +1}{\frac{1+\cos(x)}{\sin(x)} +1} \\ &\color{#D61F06} 1-\cos(x) = 2 \sin^2(x/2) \\ &\color{#D61F06} 1+\cos(x) = 2 \cos^2(x/2) \\ &\color{#D61F06} \sin(x) = 2 \sin(x/2) \cos(x/2)\\ \frac{\frac{1-\cos(x)}{\sin(x)} +1}{\frac{1+\cos(x)}{\sin(x)} +1} &= \frac{\frac{\color{#D61F06} 2 \sin^2(x/2)}{\color{#D61F06} 2 \sin(x/2) \cos(x/2)} +1}{\frac{\color{#D61F06} 2 \cos^2(x/2)}{\color{#D61F06} 2 \sin(x/2) \cos(x/2)} +1}\\ & = \frac{\tan(x/2) + 1 }{\cot(x/2)+1} \cdot \color{#D61F06} \frac{\tan(x/2)}{\tan(x/2)} \\ & = \frac{\left ( \tan(x/2) + 1 \right ) \tan(x/2) }{1+\tan(x/2)} = \tan(x/2) \end{aligned}$