## Precalculus (6th Edition) Blitzer

The value of $2\tan \frac{\pi }{3}+\cos \frac{\pi }{4}\tan \frac{\pi }{6}$ is $\frac{12\sqrt{3}+\sqrt{6}}{6}$.
Consider the given expression $2\tan \frac{\pi }{3}+\cos \frac{\pi }{4}\tan \frac{\pi }{6}$. Consider the provided right angled triangle. Now, to find $\tan \left( \frac{\pi }{3} \right)$, apply the definition of $\tan \theta$ to the provided triangle, $\tan \theta =\frac{\text{Length of side opposite of }\theta }{\text{Length of side adjacent of }\theta }$. Here, $\theta =\frac{\pi }{3}$ and $\tan \left( \frac{\pi }{3} \right)=\tan 60{}^\circ$. Therefore, \begin{align} & \tan \left( \frac{\pi }{3} \right)=\frac{\text{Length of side opposite of 60}{}^\circ }{\text{Length of side adjacent of 60}{}^\circ } \\ & =\frac{\sqrt{3}}{1} \\ & =\sqrt{3} \end{align} To find $\cos \left( \frac{\pi }{4} \right)$ apply the definition of $\cos \theta$ to the provided triangle, $\cos \theta =\frac{\text{Length of side adjacent of }\theta }{\text{hypotenuse}}$. But, here, $\theta =\frac{\pi }{4}$ , and $\cos \left( \frac{\pi }{4} \right)=\cos 45{}^\circ$. Therefore, \begin{align} & \cos \left( \frac{\pi }{4} \right)=\frac{\text{Length of side adjacent of 45}{}^\circ }{\text{Length of hypotenuse}} \\ & =\frac{1}{\sqrt{2}} \end{align} Rationalize the denominator, \begin{align} & \cos \left( \frac{\pi }{4} \right)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ & =\frac{\sqrt{2}}{2} \end{align} To find $\tan \left( \frac{\pi }{6} \right)$ apply the definition of $\tan \theta$ to the provided triangle, $\tan \theta =\frac{\text{Length of side opposite of }\theta }{\text{Length of side adjacent of }\theta }$. But, here, $\theta =\frac{\pi }{6}$ and $\tan \left( \frac{\pi }{6} \right)=\tan 30{}^\circ$. Therefore, \begin{align} & \tan \left( \frac{\pi }{6} \right)=\frac{\text{Length of side opposite of 60}{}^\circ }{\text{Length of side adjacent of 60}{}^\circ } \\ & =\frac{1}{\sqrt{3}} \end{align} Rationalize the denominator as, \begin{align} & \tan \left( \frac{\pi }{6} \right)=\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ & =\frac{\sqrt{3}}{3} \end{align} Now substitute $\tan \left( \frac{\pi }{3} \right)=\sqrt{3},\text{ cos}\left( \frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}\text{ and tan}\left( \frac{\pi }{6} \right)=\frac{\sqrt{3}}{3}$ in the provided expression, $2\tan \frac{\pi }{3}+\cos \frac{\pi }{4}\tan \frac{\pi }{6}$. Therefore, \begin{align} & 2\tan \frac{\pi }{3}+\cos \frac{\pi }{4}\tan \frac{\pi }{6}=2\left( \sqrt{3} \right)+\left( \frac{\sqrt{2}}{2} \right)\left( \frac{\sqrt{3}}{3} \right) \\ & =2\sqrt{3}+\frac{\sqrt{6}}{6} \\ & =\frac{12\sqrt{3}+\sqrt{6}}{6} \end{align} Hence the value of $2\tan \frac{\pi }{3}+\cos \frac{\pi }{4}\tan \frac{\pi }{6}$ is $\frac{12\sqrt{3}+\sqrt{6}}{6}$.