Answer
(a) $tan\frac{x}{2}=cot\frac{x}{2}-2cotx$
(b)
If $x=0$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$;
If $x\ne 0$ then $x\ne k\pi$ and $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=1/x-cotx$;
If $x=k\pi$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$; is not defined.
Work Step by Step
(a) ${cotx/2-tanx/2}=\frac{cosx/2}{sinx/2}-\frac{sinx/2}{cosx/2}$
$=2cotx$
Hence, $tan\frac{x}{2}=cot\frac{x}{2}-2cotx$
(b)
If $x=0$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$;
If $x\ne 0$ then $x\ne k\pi$ and $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=1/x-cotx$;
If $x=k\pi$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$; is not defined.