Reciprocal of Difference of Squares as Difference of Reciprocals
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Theorem
- $\dfrac 1 {x^2 - y^2} = \dfrac 1 {2 y \paren {x - y} } - \dfrac 1 {2 y \paren {x + y} }$
Proof
\(\ds \dfrac 1 {x - y} - \dfrac 1 {x + y}\) | \(=\) | \(\ds \dfrac {\paren {x + y} - \paren {x - y} } {\paren {x + y} \paren {x - y} }\) | putting everything over a common denominator | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 y} {x^2 - y^2}\) | simplifying, and Difference of Two Squares | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 {2 y \paren {x - y} } - \dfrac 1 {2 y \paren {x + y} }\) | \(=\) | \(\ds \dfrac 1 {x^2 - y^2}\) | dividing both sides by $2 y$ |
$\blacksquare$