Prove $\sum_{n\geq1}\frac{2^n (1-\cos\frac{x}{2^n})^2}{\sin\frac{x}{2^{n-1}}}=\tan\frac{x}{2}-\frac{x}{2}$

Here is a geometric proof for

$$\sum_{n\geq1}\frac{2(1-\cos\frac{x}{2^n})}{\sin\frac{x}{2^{n-1}}}=\tan\frac{x}{2}$$

Let $OA_1,\>OA_2,\>OA_3 …$ successively bisect the vextex angle $\angle O=\frac x2$ of the right triangle $OTA$, which leads to $\frac{AA_1}{TA_1} = \frac{OA}{OT}$, or,

$$AA_1= OA\cdot TA_1 = \sec\frac x2\tan\frac x4 = \frac{2\sin^2\frac x4}{\cos\frac x2\sin\frac x2} = \frac{2(1-\cos\frac x2)}{\sin x}$$

and, likewise,

$$A_1A_2 = \frac{2(1-\cos\frac x4)}{\sin \frac x2},\>\>\>\>\> A_2A_3 = \frac{2(1-\cos\frac x8)}{\sin \frac x4},\>\>\>\>\>A_3A_4 = ...$$

From the diagram, we have

$$\tan \frac x2 = AT = AA_1 + A_1A_2 + A_2A_3 + … = \sum_{n\geq1}\frac{2(1-\cos\frac{x}{2^n})}{\sin\frac{x}{2^{n-1}}}$$