[tex]$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}=\frac{b-a}{b+a}[/tex]
Solution:
Given expression:
[tex]$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}[/tex]
To simplify the given expression:
[tex]$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}[/tex]
Let us first solve the numerator of the expression [tex]\frac{1}{a} -\frac{1}{b}[/tex].
To add or subtract the fraction, the denominators of the fraction must be same.
To make it same, multiply [tex]\frac{1}{a}[/tex] by [tex]\frac{b}{b}[/tex] and [tex]\frac{1}{b}[/tex] by [tex]\frac{a}{a}[/tex].
[tex]$\frac{1}{a} -\frac{1}{b}=\frac{b}{ab} -\frac{a}{ab}[/tex]
[tex]$=\frac{b-a}{ab}[/tex] ------------------- (1)
Now, solve the denominator of the expression [tex]\frac{1}{a} +\frac{1}{b}[/tex].
To add or subtract the fraction, the denominators of the fraction must be same.
To make it same, multiply [tex]\frac{1}{a}[/tex] by [tex]\frac{b}{b}[/tex] and [tex]\frac{1}{b}[/tex] by [tex]\frac{a}{a}[/tex].
[tex]$\frac{1}{a} +\frac{1}{b}=\frac{b}{ab} +\frac{a}{ab}[/tex]
[tex]$=\frac{b+a}{ab}[/tex] ------------------- (2)
Substitute (1) and (2) in the given expression.
[tex]$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}=\frac{\frac{b-a}{ab}}{\frac{b+a}{ab}}[/tex]
Using rational rule: [tex]$\frac{\frac{x}{y} }{\frac{w}{z} }=\frac{x}{y} \times\frac{z}{w}[/tex]
[tex]$=\frac{b-a}{ab}}\times {\frac{ab}{b+a}}[/tex]
Common factor ab get canceled.
[tex]$=\frac{b-a}{b+a}[/tex]
[tex]$\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}=\frac{b-a}{b+a}[/tex]
Hence the simplified expression is [tex]\frac{b-a}{b+a}[/tex].
