Establish the identity: $$\frac{\sin(\alpha+\beta)}{\cos\alpha \cos\beta} = \tan\alpha+\tan\beta$$

I'm having a hard time solving this one. I changed the numerator to $\sin\alpha\cos\beta+\cos\alpha\sin\beta$ and the denominator to $\frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)]$, but I'm not really sure what to do from there. Multiplying by the conjugate didn't seem to get me anywhere, either. I'd really appreciate help with this!


You're working too hard. Just change the numerator (not the denominator), break the fraction into two, and simplify.

  • $\begingroup$ Thank you! I always overthink this stuff. I didn't expect it to be that simple, it never occurred to me to break it up. $\endgroup$ – jorsully Nov 29 '16 at 5:43


$$ \Leftrightarrow \frac{\sin(a)\cos(b)+\cos(a)\sin(b)}{\cos(a)\cos(b)} $$

$$ \Leftrightarrow \frac{\sin(a)\cos(b)}{\cos(a)\cos(b)} +\frac{\cos(a)\sin(b)}{\cos(a)\cos(b)} $$

$$ \therefore \tan(a) + \tan(b) $$


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