The best I can tell, you are essentially given to verify:
\(\displaystyle \dfrac{A\sin(\theta)}{A+A\cos(\theta)}= \tan\left(\dfrac{\theta}{2}\right)\)
I would first divide the numerator and denominator of the LHS by \(\displaystyle A\) to get:
\(\displaystyle \dfrac{\sin(\theta)}{1+\cos(\theta)}= \tan\left(\dfrac{\theta}{2}\right)\)
The double-angle identity for sine is \(\displaystyle \sin(2\alpha)=2\sin(\alpha)\cos(\alpha)\) which means:
\(\displaystyle \sin(\theta)=2\sin\left(\dfrac{\theta}{2}\right) \cos\left(\dfrac{\theta}{2}\right)\)
And in this thread, we've already established that:
\(\displaystyle 1+\cos(\theta)= 2\cos^2\left(\dfrac{\theta}{2}\right)\)
And so applying these to the identity we are given to verify, we obtain:
\(\displaystyle \dfrac{2\sin\left( \dfrac{\theta}{2}\right) \cos\left(\dfrac{ \theta}{2}\right)}{ 2\cos^2\left(\dfrac{ \theta}{2}\right)}= \tan\left(\dfrac{\theta}{2}\right)\)
Divide the numerator and denominator of the LHS by \(\displaystyle 2\cos\left(\dfrac{\theta}{2}\right)\) to get:
\(\displaystyle \dfrac{ \sin\left(\dfrac{\theta}{2}\right)}{ \cos\left(\dfrac{\theta}{2}\right)}= \tan\left(\dfrac{\theta}{2}\right)\)
Using the identity \(\displaystyle \dfrac{\sin(\alpha)}{\cos(\alpha)}=\tan(\alpha)\) we have:
\(\displaystyle \tan\left(\dfrac{\theta}{2}\right)= \tan\left(\dfrac{\theta}{2}\right) \quad\checkmark\)
Shown as desired.
In the future, I suggest beginning a new thread for a new question. This way threads don't become potentially convoluted and hard to follow.