## Trigonometry (11th Edition) Clone

$$\frac{\tan(\alpha+\beta)-\tan\beta}{1+\tan(\alpha+\beta)\tan\beta}=\tan\alpha$$ The equation is verified to be an identity.
$$\frac{\tan(\alpha+\beta)-\tan\beta}{1+\tan(\alpha+\beta)\tan\beta}=\tan\alpha$$ If we look closely, we might realize that the left side is actually the result of an identity involving the difference of tangent of $(\alpha+\beta)$ and $\beta$. In other words,$$\frac{\tan(\alpha+\beta)-\tan\beta}{1+\tan(\alpha+\beta)\tan\beta}=\tan[(\alpha+\beta)-\beta]=\tan\alpha$$ The equation is therefore verified to be an identity.