Answer
The value of $2\tan \frac{\pi }{3}+\cos \frac{\pi }{4}\tan \frac{\pi }{6}$ is $\frac{12\sqrt{3}+\sqrt{6}}{6}$.
Work Step by Step
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 $\frac{12\sqrt{3}+\sqrt{6}}{6}$.
