How do I simplify $\tan(\alpha-\beta)$ into $\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$?
How do I simplify $\tan(\alpha-\beta)$ into $\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$? I tried: $$\tan(\alpha-\beta) = \\\frac{\sin(\alpha-\bet...