buktikan bahwa sin(a-b)/sin(a+b) = tan a - tan b/ tan a +tan b

[tex]\displaystyle \displaystyle \frac{\sin(a-b)}{\sin(a+b)}=\frac{\tan a-\tan b}{\tan a+\tan b}\\\frac{\sin a\cos b-\cos a\sin b}{\sin a\cos b+\cos a\sin b}=\frac{\tan a-\tan b}{\tan a+\tan b}\\\frac{\sin a\cos b-\cos a\sin b}{\sin a\cos b+\cos a\sin b}\cdot\frac{\frac{1}{\cos a\cos b}}{\frac{1}{\cos a\cos b}}=\frac{\tan a-\tan b}{\tan a+\tan b}\\\frac{\frac{\sin a\cos b-\cos a\sin b}{\cos a\cos b}}{\frac{\sin a\cos b+\cos a\sin b}{\cos a\cos b}}=\frac{\tan a-\tan b}{\tan a+\tan b}[/tex]
[tex]\displaystyle \frac{\frac{\sin a}{\cos a}-\frac{\sin b}{\cos b}}{\frac{\sin a}{\cos a}+\frac{\sin b}{\cos b}}=\frac{\tan a-\tan b}{\tan a+\tan b}\\\boxed{\boxed{\frac{\tan a-\tan b}{\tan a+\tan b}=\frac{\tan a-\tan b}{\tan a+\tan b}}}[/tex]