Remember that
[tex]\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a\cdot\tan b}[/tex]
In this problem
a=5pi/12
b=pi/4
substitute
[tex]\begin{gathered} \tan (\frac{5\pi}{12}-\frac{\pi}{4})=\frac{\tan \frac{5\pi}{12}-\tan \frac{\pi}{4}}{1+\tan \frac{5\pi}{12}\cdot\tan \frac{\pi}{4}} \\ \\ \tan (\frac{\pi}{6})=\tan (\frac{5\pi}{12}-\frac{\pi}{4})=\frac{\tan\frac{5\pi}{12}-\tan\frac{\pi}{4}}{1+\tan\frac{5\pi}{12}\cdot\tan\frac{\pi}{4}} \end{gathered}[/tex]
The answer is
[tex]\tan (\frac{\pi}{6})[/tex]
