Chapter 7 - Analytic Trigonometry - 7.5 Sum and Difference Formulas - 7.5 Assess Your Understanding - Page 488: 55

Since $\tan(\alpha-\beta)=\dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$, then $\tan(\pi-\theta)\\ =\dfrac{\tan(\pi)-\tan(\theta)}{1+\tan(\pi)\tan(\theta)}\\ =\dfrac{0-\tan(\theta)}{1+0\tan(\theta)}\\ =\dfrac{-\tan(\theta)}{1}\\ =-\tan(\theta)$

Recall: $\tan(\alpha-\beta)=\dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$ Hence, $\tan(\pi-\theta)\\ =\dfrac{\tan(\pi)-\tan(\theta)}{1+\tan(\pi)\tan(\theta)}\\ =\dfrac{0-\tan(\theta)}{1+0\tan(\theta)}\\ =\dfrac{-\tan(\theta)}{1}\\ =-\tan(\theta)$