use the appropriate angle sum or difference formula to find the exact value of $\\tan(-\\frac{17\\pi}{12})$.

Explanation:

Step1: Use the property of tangent function

Since (\tan(-\alpha)=-\tan\alpha), then (\tan\left(-\frac{17\pi}{12}\right)=-\tan\left(\frac{17\pi}{12}\right)). And (\frac{17\pi}{12}=\frac{9\pi}{12}+\frac{8\pi}{12}=\frac{3\pi}{4}+\frac{2\pi}{3}).

Step2: Apply the tangent sum formula

The tangent sum formula is (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}). Here (A=\frac{3\pi}{4}), (\tan A=- 1); (B = \frac{2\pi}{3}), (\tan B=-\sqrt{3}). [ \begin{align*} \tan\left(\frac{3\pi}{4}+\frac{2\pi}{3}\right)&=\frac{\tan\frac{3\pi}{4}+\tan\frac{2\pi}{3}}{1-\tan\frac{3\pi}{4}\tan\frac{2\pi}{3}}\ &=\frac{-1-\sqrt{3}}{1-(-1)\times(-\sqrt{3})}\ &=\frac{-1 - \sqrt{3}}{1-\sqrt{3}} \end{align*} ]

Step3: Rationalize the denominator

Multiply the numerator and denominator by (1 + \sqrt{3}) [ \begin{align*} \frac{-1-\sqrt{3}}{1-\sqrt{3}}\times\frac{1+\sqrt{3}}{1+\sqrt{3}}&=\frac{(-1-\sqrt{3})(1+\sqrt{3})}{1-3}\ &=\frac{-1-\sqrt{3}-\sqrt{3}-3}{-2}\ &=\frac{-4 - 2\sqrt{3}}{-2}\ &=2+\sqrt{3} \end{align*} ] Since (\tan\left(-\frac{17\pi}{12}\right)=-\tan\left(\frac{17\pi}{12}\right))

Answer:

(2+\sqrt{3})