If $\tan x=p+1$ and $\tan y=p-1$, then the value of $2 \cot (x-y)$ is
Answer: $\mathbf B$
Explanation: We know that: $\begin {align} \tan\mathrm{(x-y)} &= \mathrm{\frac{\tan x - \tan y}{1 + \tan x \tan y}}\\\\ &= \mathrm{\frac{p + 1 - (p -1)} {1 + (p+1)(p-1)}}\\\\ &= \mathrm{\frac{2}{1 + p^2 - 1^2}}\\\\ &= \mathrm{\frac{2}{p^2}}\end{align}$ $\therefore \cot \mathrm{(x-y)} =\mathrm{\frac{1}{\tan (x-y)}=\frac{1}{\dfrac{2}{p^2}}= \frac{p^2}{2}}$ $\therefore \mathrm{2\times\cot(x-y) \require {cancel}= \frac{p^2}{\cancel{2}}\times \cancel[red]{2} = p^2}$ $\therefore \mathbf B$ is the correct option.