To find the exact value of \(\tan\left(\frac{11\pi}{12}\right)\), we can use the angle addition formula for tangent. We can express \(\frac{11\pi}{12}\) in terms of angles that are more convenient to work with.
Notice that: \[ \frac{11\pi}{12} = \frac{6\pi}{12} + \frac{5\pi}{12} = \frac{\pi}{2} + \frac{5\pi}{12} \] However, it is better to express it as: \[ \frac{11\pi}{12} = \pi - \frac{\pi}{12} \]
Using the tangent identity \(\tan(\pi - x) = -\tan(x)\), we have: \[ \tan\left(\frac{11\pi}{12}\right) = \tan\left(\pi - \frac{\pi}{12}\right) = -\tan\left(\frac{\pi}{12}\right) \]
Now we need to find \(\tan\left(\frac{\pi}{12}\right)\). We know that: \[ \frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4} \] Using the tangent difference formula: \[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \] where \(a = \frac{\pi}{3}\) and \(b = \frac{\pi}{4}\). We have: \[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3}, \quad \tan\left(\frac{\pi}{4}\right) = 1 \]
Applying the formula, we get: \[ \tan\left(\frac{\pi}{12}\right) = \frac{\tan\left(\frac{\pi}{3}\right) - \tan\left(\frac{\pi}{4}\right)}{1 + \tan\left(\frac{\pi}{3}\right) \tan\left(\frac{\pi}{4}\right)} = \frac{\sqrt{3} - 1}{1 + \sqrt{3} \cdot 1} \]
Thus, this simplifies to: \[ \tan\left(\frac{\pi}{12}\right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} \]
Next, we can rationalize the denominator: \[ \tan\left(\frac{\pi}{12}\right) = \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{3 - \sqrt{3} - \sqrt{3} + 1}{1 - 3} = \frac{4 - 2\sqrt{3}}{-2} = \frac{2 - \sqrt{3}}{-1} = \sqrt{3} - 2 \]
Thus, \[ \tan\left(\frac{11\pi}{12}\right) = -\tan\left(\frac{\pi}{12}\right) = -( \sqrt{3} - 2 ) = 2 - \sqrt{3} \]
So the exact value of \(\tan\left(\frac{11\pi}{12}\right)\) is: \[ \boxed{2 - \sqrt{3}} \]