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Mathematics

If p+1p=x and p−1p=yp + \dfrac{1}{p} = x\text{ and }p - \dfrac{1}{p} = y, then the relation between x and y is :

  1. x2 = y2

  2. x2 + y2 = 4

  3. x2 - y2 = 4

  4. xy = 2

Expansions

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Answer

Given,

Upon squaring p+1p=xp + \dfrac{1}{p} = x we get,

⇒(p+1p)2=x2⇒(p2+1p2+2)=x2 ….(1)\Rightarrow \Big(p + \dfrac{1}{p}\Big)^2 = x^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2\Big) = x^2 \text{ ….(1)}

Upon squaring p−1p=yp - \dfrac{1}{p} = y we get,

⇒(p−1p)2=y2⇒(p2+1p2−2)=y2 ….(2)\Rightarrow \Big(p - \dfrac{1}{p}\Big)^2 = y^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} - 2\Big) = y^2 \text{ ….(2)}

Subtracting (2) from (1) we get,

⇒(p2+1p2+2)−(p2+1p2−2)=x2−y2⇒(p2+1p2+2−p2−1p2+2)=x2−y2⇒x2−y2=4.\Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2\Big) - \Big(p^2 + \dfrac{1}{p^2} - 2\Big) = x^2 - y^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2 - p^2 - \dfrac{1}{p^2} + 2\Big) = x^2 - y^2 \\[1em] \Rightarrow x^2 - y^2 = 4.

Hence, Option 3 is the correct option.

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