Simplify a rational identity 
Simplify:
  $$\frac{\dfrac{a}{b}-\dfrac{b}{a}}{1+\dfrac{b}{a}}$$

I have a feeling the solution has to do with factoring, but I'm really not sure, and would appreciate any help.
 A: Multiply with $\frac{ab}{ab}$ as first step:
$$\frac{\frac ab-\frac ba}{1+\frac ba}=\frac{a^2-b^2}{ab+b^2}=\frac{(a+b)(a-b)}{(a+b)b}=\frac{a-b}b=\frac ab-1 $$
(provided $a+b\ne0$, but in that case the original fraction would not be defined)
A: Multiply numerator and denominator by $ab$ to get $\frac {a^{2}-b^{2}} {b(a+b)}$. Then use the fact that $a^{2}-b^{2} =(a-b)(a+b)$. Cancel $a+b$ to get $\frac {a-b} b$.
A: Well, in view of addition the rule is 
$$\frac{a}{b}+\frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad+bc}{bd}$$
i.e., the fractions need to be extended first to obtain the same denominator and then the addition can be performed (same for subtraction).
In view of multiplication the rule is
$$\frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}$$
i.e., numerator and denominator are multiplied separately.
In view of division the rule is
$$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b}:\frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}$$
i.e., in the fraction (denominator) $c/d$ the numerator and denominator are flipped $d/c$ and then multiplied with $a/b$.
You can easily use these rules to simplify your fraction:
$$\frac{\frac{a}{b}-\frac{b}{a}}{1+\frac{b}{a}} 
= \frac{\frac{a^2}{ab}-\frac{b^2}{ab}}{\frac{a}{a}+\frac{b}{a}}
= \frac{\frac{a^2-b^2}{ab}}{\frac{a+b}{a}}
= \frac{a^2-b^2}{ab}\cdot \frac{a}{a+b}
= \frac{(a^2-b^2)a}{ab(a+b)}
= \frac{a^2-b^2}{b(a+b)}
=\frac{(a-b)(a+b)}{b(a+b)}
=\frac{a-b}{b}.$$
