Answer
(a) $tan\frac{x}{2}=cot\frac{x}{2}-2cotx$
(b) (1)When $x=0$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$
(2) when $x\ne 0$ then $x\ne k\pi$ and $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=1/x-costx$
(3) When $x=k\pi$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$; which is undefined.
Work Step by Step
(a) Since,$cot\frac{x}{2}=\dfrac{cos\frac{x}{2}}{sin\frac{x}{2}}$
and
$tan\frac{x}{2}=\dfrac{sin\frac{x}{2}}{cos\frac{x}{2}}$
As per question: ${cot\frac{x}{2}-tan\frac{x}{2}}=\dfrac{cos\frac{x}{2}}{sin\frac{x}{2}}-\dfrac{sin\frac{x}{2}}{cos\frac{x}{2}}=2cotx$
or, $tan\frac{x}{2}=cot\frac{x}{2}-2cotx$
(b) (1)When $x=0$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$
(2) when $x\ne 0$ then $x\ne k\pi$ and $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=1/x-costx$
(3) When $x=k\pi$ then $\Sigma_{n=1}^{\infty}\frac{1}{2^n}tan\frac{x}{2^n}=0$; which is undefined.
