Answer
$tan\frac{11\pi}{12} = -\frac{1}{2+\sqrt3} \approx -0.26795$
Work Step by Step
We know,
$$tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$$
&
$$tan(\pi-A)= -tanA$$
Using above two formulas we can write,
$tan\frac{11\pi}{12}= tan(\pi-\frac{\pi}{12})=-tan\frac{\pi}{12}$
Now we can write,
$\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$
So we use that in above equation,
$$tan\frac{\pi}{12}=\frac{tan\frac{\pi}{3}-tan\frac{\pi}{4}}{1+tan\frac{\pi}{3}tan\frac{\pi}{4}}$$
putting the values of respective trigonometric identities
$$tan\frac{\pi}{12}=\frac{\sqrt3-1}{1+(\sqrt3\times1)}$$
$tan\frac{\pi}{12}=\frac{\sqrt3-1}{1+\sqrt3}=\frac{\sqrt3-1}{1+\sqrt3}\times(\frac{1+\sqrt3}{1+\sqrt3})$
$tan\frac{\pi}{12}=\frac{3-1}{1+3+2\sqrt3}=\frac{2}{4+2\sqrt3}=\frac{1}{2+\sqrt3}$
So we get,
$-tan\frac{\pi}{12}=-\frac{1}{2+\sqrt3}$
$$\therefore tan\frac{11\pi}{12}=-\frac{1}{2+\sqrt3}\approx-0.26795$$
