Given that \begin{align*} \sin(\theta + \phi) &= \sin\theta\cos\phi + \cos\theta\sin \phi\\ \cos(\theta + \phi) &= \cos\theta\cos\phi - \sin\theta\sin\phi \end{align*} find $\tan(\theta + \phi)$ in terms of $\tan\theta$ and $\tan\phi$. Show hint

$\tan x = \sin x/\cos x$

Firstly we need to write tan in terms of sin and cos so we can apply the formulae we already know: \begin{align*} \tan(\theta + \phi) &= \frac{\sin(\theta + \phi)}{\cos(\theta + \phi)}\\ &= \frac{\sin\theta\cos\phi + \cos\theta\sin\phi}{\cos\theta\cos\phi - \sin\theta\sin\phi}. \end{align*} Now we want everything in terms of $\tan\theta$ and $\tan\phi$. It’s probably not obvious how to do this at first. Noticing the $\sin\theta$ in the first term of the numerator, we’d like to divide that by $\cos\theta$ so it becomes $\tan\theta$. And similarly, since the second term contains $\sin\phi$, we’d like to divide that by $\cos\phi$. So a good idea might be to divide both top and bottom of the fraction by $\cos\theta\cos\phi$. Doing this yields: \[ \tan(\theta + \phi) = \frac{\tan\theta + \tan\phi}{1 - \tan\theta\tan\phi} \]