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The addition formula for tangent will be achieved via brute force from the addition formulas for sine and cosine.
$\begin{array}{rl} \tan(\alpha+\beta) & =\displaystyle \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} \\ & \\ & =\displaystyle \frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta}{\cos \alpha \cos \beta -\sin \alpha \sin \beta} \\ & \\ & =\displaystyle \frac{ \displaystyle \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\cos \beta}{\cos \beta}+\displaystyle \frac{\cos \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta}}{ \displaystyle \frac{\cos \alpha}{\cos \alpha} \cdot \frac{\cos \beta}{\cos \beta}-\frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \beta}{\cos \beta}} \\ & \\ & =\displaystyle \frac{ \tan \alpha \cdot 1+1 \cdot \tan \beta}{1 \cdot 1-\tan \alpha \tan \beta} \\ & \\ & =\displaystyle \frac{ \tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \end{array}$
Note that $\tan$ is an odd function, i.e. $\tan(-x)=-\tan x$ This fact enables us to obtain the subtraction formula for tangent.
$$\tan(\alpha-\beta)=\tan(\alpha+(-\beta))=\frac{ \tan \alpha+\tan(-\beta)}{1-\tan \alpha \tan(-\beta)}=\frac{ \tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}$$
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