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

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!

## 2 Answers

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

• 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. – jorsully Nov 29 '16 at 5:43

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

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

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

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