2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

From

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.