Answer:
The proposition is true.
Step-by-step explanation:
Now we proceed to demostrate that expression given is true by algebraic means:
1) [tex]\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1} }{y^{-1}}[/tex] Given
2) [tex]\frac{y^{-1}\cdot (x^{-1}+y^{-1})+x^{-1}\cdot (x^{-1}-y^{-1})}{x^{-1}\cdot y^{-1}}[/tex] [tex]\frac{a}{b} + \frac{c}{d} = \frac{a\cdot d+b\cdot c}{b\cdot d}[/tex]
3) [tex]\frac{x^{-1}\cdot y^{-1}+y^{-2}+x^{-2}-x^{-1}\cdot y^{-1}}{x^{-1}\cdot y^{-1}}[/tex] Distributive property/[tex]a^{b}\cdot a^{c} = a^{b+c}[/tex]
4) [tex]\frac{x^{-2}+y^{-2}}{(x\cdot y)^{-1}}[/tex] Commutative, associative and modulative properties/Existence of additive inverse/[tex]a^{b}\cdot c^{b} = (a\cdot c)^{b}[/tex]
5) [tex][(x\cdot y)^{-1}]^{-1}\cdot (x^{-2}+y^{-2})[/tex] Commutative property/Definition of division
6) [tex](x\cdot y)\cdot (x^{-2}+y^{-2})[/tex] [tex](x^{-1})^{-1}[/tex]
7) [tex]x^{-1}\cdot y + x\cdot y^{-1}[/tex] Distributive property/Associative property/[tex]a^{b}\cdot a^{c} = a^{b+c}[/tex]
8) [tex](x^{-1}\cdot y^{-1})\cdot y^{2} + (x^{-1}\cdot y^{-1})\cdot x^{2}[/tex] Modulative property/Existence of additive inverse/[tex]a^{b}\cdot a^{c} = a^{b+c}[/tex]
9) [tex](x^{-1}\cdot y^{-1})\cdot (x^{2}+y^{2})[/tex] Distributive property
10) [tex](x\cdot y)^{-1}\cdot (x^{2}+y^{2})[/tex] [tex]a^{b}\cdot c^{b} = (a\cdot c)^{b}[/tex]
11) [tex]\frac{x^{2}+y^{2}}{x\cdot y}[/tex] Commutative property/Definition of division/Result
