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$$\begin{align}2\cot{(x-y)}&=\frac{2}{\tan{(x-y)}}=\frac{2}{\frac{\tan x - \tan y}{1+\tan x \tan y}}\\&=\frac{2(1+\tan x \tan y)}{\tan x - \tan y}\\&=\frac{2\{1+(p+1)(p-1)\}}{p+1-(p-1)}\\&=\frac{2\{1+p^2-1\}}{2}\\&=p^2\end{align}$$

 

So the correct answer is B.

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$cot(x-y) = \frac{cotx.coty+1}{coty-cotx}$. Given $tanx = p+1$ and $tany= p-1$. So, $cotx = \frac{1}{p+1}$ and $coty = \frac{1}{p-1}$.

$cot(x-y)= \frac{cotx.coty+1}{coty-cotx} = \frac{\frac{p^2}{p^2-1}}{\frac{2}{p^2-1}}=\frac{p^2}{2}$

So, $2cot(x-y) = p^2$ (Option B)
Answer:

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