If cot A = [sin B/(1 - cos B)], then what is the value of cot 2A?

Given : $$cotA=\frac{sinB}{1-cosB}$$ -----------(i)

To find : $$cot2A=\frac{cot^2A-1}{2cotA}$$

Substituting value from equation (i),

= $$[(\frac{sinB}{1-cosB})^2-1]\div[2(\frac{sinB}{1-cosB})]$$

= $$(\frac{sin^2B-(1-cosB)^2}{(1-cosB)^2})\times(\frac{(1-cosB)}{2sinB})$$

= $$\frac{sin^2B-(1-cosB)^2}{2sinB(1-cosB)}$$

= $$\frac{sin^2B-(1+cos^2B-2cosB)}{2sinB(1-cosB)}$$

= $$\frac{sin^2B-1-cos^2B+2cosB}{2sinB(1-cosB)}$$

Using, $$sin^2\theta-1=-cos^2\theta$$

= $$\frac{-2cos^2B+2cosB}{2sinB(1-cosB)}$$

= $$\frac{2cosB(1-cosB)}{2sinB(1-cosB)}$$

= $$\frac{cosB}{sinB}=cotB$$

=> Ans - (C)