Chapter 6 - Analytic Trigonometry - Section 6.5 Sum and Difference Formulas - 6.5 Assess Your Understanding - Page 508: 28

$$\dfrac{\sqrt 3}{3}$$

Apply the difference formula: $\tan(A -B)=\dfrac{\tan A - \tan B}{1 +\tan A \ \tan B }$ Therefore, $$ \dfrac{\tan 40^\circ - \tan 10^\circ}{1-\tan 40^\circ \ \tan 10^\circ } =\tan (40^\circ -10^\circ) \\ =\tan \ (30^\circ) \\ =\dfrac{1}{\sqrt 3} \times \dfrac{\sqrt 3}{\sqrt 3} \\ =\dfrac{\sqrt 3}{3}$$