How do I simplify $\tan(\alpha-\beta)$ into $\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$?
I tried:
$$\tan(\alpha-\beta) = \\\frac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)}=\\\frac{\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)} = \\\frac{\sin\alpha\cos\beta}{\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)}-\frac{\cos\alpha\sin\beta}{\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)} = ???$$
What do I do next?
$\endgroup$ 41 Answer
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$$ \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} $$
divide the numerator and denominator by $\cos \alpha \cos \beta$.
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