Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.

Prove that,
  $$
\dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2}
$$


Using the identities$$1+\cos A=2 \cos ^{2} \frac{A}{2}, 1-\sin A=2 \sin ^{2} \frac{1}{2} , \sin A=2 \sin\frac{\theta}{2} \cos\frac{\theta}{2},
$$
We have $$
\begin{aligned}
\frac{1+\sin A-\cos A}{1+\sin A+\cos A} &=\frac{2 \sin ^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos ^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}} \\
&=\frac{\sin \frac{A}{2}\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)}{\cot \frac{A}{2}\left(\cos \frac{A}{2}+\sin \frac{A}{2}\right)} \\
&=\tan \frac{A}{2}
\end{aligned}
$$
