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?

  • $\begingroup$ You don't need to split the fraction as you did, you can get the result from the previous line. $\endgroup$ – John Doe Apr 26 '17 at 16:31
  • $\begingroup$ You can also work backwards to figure out the steps... $\endgroup$ – farruhota May 4 '17 at 9:33
  • $\begingroup$ @FarrukhAtaev How do I do that? $\endgroup$ – Mark Read May 4 '17 at 14:58
  • $\begingroup$ $\frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}=\frac{{\frac{\sin{\alpha}}{\cos{\alpha}}}-\frac{\sin{\beta}}{\cos{\beta}}}{{1}+\frac{\sin{\alpha}}{\cos{\alpha}}\cdot\frac{\sin{\beta}}{\cos{\beta}}}=\frac{\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}}=\frac{\sin{(\alpha-\beta})}{\cos{(\alpha-\beta})}=\tan{(\alpha-\beta)}$ $\endgroup$ – farruhota May 5 '17 at 10:17


$$ \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} $$

divide the numerator and denominator by $\cos \alpha \cos \beta$.


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