Precalculus (6th Edition) Blitzer

The exact value of $\tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)$ is $2+\sqrt{3}$.
Use the sum formula of tangent and evaluate the expression as, $\tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\frac{\tan \frac{\pi }{6}+\tan \frac{\pi }{4}}{1-\tan \frac{\pi }{6}\tan \frac{\pi }{4}}$ Substitute the values, $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$ and $\tan \frac{\pi }{4}=1$. \begin{align} & \tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\frac{\frac{1}{\sqrt{3}}+1}{1-\left( \frac{1}{\sqrt{3}}\times 1 \right)} \\ & =\frac{1+\sqrt{3}}{\sqrt{3}-1} \end{align} Further, rationalize the result and solve as, \begin{align} & \tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=-\left( \frac{1+\sqrt{3}}{1-\sqrt{3}} \right)\times \frac{1+\sqrt{3}}{1+\sqrt{3}} \\ & =-\frac{{{\left( 1+\sqrt{3} \right)}^{2}}}{{{1}^{2}}-{{\left( \sqrt{3} \right)}^{2}}} \\ & =-\frac{\left( 1+3+2\sqrt{3} \right)}{1-3} \\ & =\frac{4+2\sqrt{3}}{2} \end{align} Thus, \begin{align} & \tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\frac{2\left( 2+\sqrt{3} \right)}{2} \\ & =2+\sqrt{3} \end{align} Hence, the exact value of $\tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)$ is equivalent to $2+\sqrt{3}$.