## Trigonometry (11th Edition) Clone

$\frac{a-b}{a+b} = \frac{sin~A-sin~B}{sin~A+sin~B}$
We can use the law of sines to find an expression for $a$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $a = \frac{b~sin~A}{sin~B}$ We can prove the statement in the question: $\frac{a-b}{a+b} = \frac{\frac{b~sin~A}{sin~B}-b}{\frac{b~sin~A}{sin~B}+b}$ $\frac{a-b}{a+b} = \frac{b~(\frac{sin~A}{sin~B}-1)}{b~(\frac{sin~A}{sin~B}+1)}$ $\frac{a-b}{a+b} = \frac{\frac{sin~A}{sin~B}-1}{\frac{sin~A}{sin~B}+1}$ $\frac{a-b}{a+b} = \frac{\frac{sin~A}{sin~B}-\frac{sin~B}{sin~B}}{\frac{sin~A}{sin~B}+\frac{sin~B}{sin~B}}$ $\frac{a-b}{a+b} = \frac{\frac{sin~A-sin~B}{sin~B}}{\frac{sin~A+sin~B}{sin~B}}$ $\frac{a-b}{a+b} = \frac{sin~A-sin~B}{sin~A+sin~B}$