Use the sum formula for the tangent function

\tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta }{1-\tan \alpha\tan \beta}

\tan \dfrac{11\pi}{12} = \tan(\dfrac{2\pi}{3}+\dfrac{\pi }{4}) = \dfrac{\tan \dfrac{2\pi}{3} + \tan \dfrac{\pi }{4} }{1-\tan \dfrac{2\pi}{3}\tan \dfrac{\pi }{4}}

=\dfrac{1-\sqrt{3} }{1+\sqrt{3} } = \dfrac{(1-\sqrt{3} )^2}{1-3} = -2+\sqrt{3}