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If $\tan x=p+1$ and $\tan y=p-1$, then the value of $2 \cot (x-y)$ is

  1. $2p$
  2. $p^2$
  3. $(p+1)(p-1)$
  4. $\frac{2p}{p^2-1}$
in Numerical Ability by Boss (16.8k points)
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2 Answers

+2 votes

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.

by Boss (14k points)
edited by
+1 vote
tan(x-y)=tanx -tany/1+tanxtany=p+1-p+1/1-(p2-1)=2/p2;

2cot(x-y)=2/tan(x-y)=2*p2/2=p2

option A
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