# Proving the tangent addition identity (by multiplying the numerator and denominator by $\cos \theta \cos \phi$ to simplify in terms of $\tan$)

In the tangent addition identity, $$\begin{eqnarray} \tan(\theta + \phi) &=& \frac{\sin(\theta + \phi)}{\cos(\theta + \phi)} \\\\ &=& \frac{\sin\theta\cos\phi + \cos\theta\sin\phi}{\cos\theta\cos\phi - \sin\theta\sin\phi} \\\\ &=& \frac{\left(\frac{\sin\theta\cos\phi + \cos\theta\sin\phi}{\cos\theta\cos\phi}\right)}{\left(\frac{\cos\theta\cos\phi - \sin\theta\sin\phi}{\cos\theta\cos\phi}\right)} \\\\ &=& \frac{\left(\frac{\sin\theta}{\cos\theta} + \frac{\sin\phi}{\cos\phi}\right)}{\left(1-\frac{\sin\theta\sin\phi}{\cos\theta\cos\phi}\right)} \\\\ &=& \frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi} \\\\ \end{eqnarray}$$ I understand how this works, but I'm trying to understand why we multiply numerator and denominator in the third line by $$\frac{1}{\cos\theta\cos\phi}$$ . I get that this allows us to simplify the solution in terms of $$\tan, \theta,$$ and $$\phi$$ (or whatever variable is used for the proof), but is that the only reason we further simplify it beyond that step?

• I mean, why not?
– anon
Commented Jun 11, 2021 at 23:26
• In a word, yes, the reason is to express $\tan(A+B)$ in terms of $\tan A$ and $\tan B.$ It’s nicer that way. To express it in terms of the bad old days, you only have to use one lookup table. (Before computers and calculators, we had tables of logarithms and trig functions. We could have a table for values of $\tan A$ for $A=1^\circ,2^\circ,\dots,360^\circ$ and then tables for smaller $B$ for values $B=0.01,0.02,\dots,0.99$ and compute a value for $\tan(A+B)$ if you needed that fineness. Commented Jun 11, 2021 at 23:48

For tangent, you can keep a table for $$\theta=0^o,1^o,\dots,89^o$$ and a table for $$\theta=0.01,\dots,0.99$$ then use this formula for $$\tan(A+B)$$ to get finer angles by picking $$A$$ from the first table and $$B$$ from the second.
The other functions would require four tables to compute, say $$\sin(A+B).$$