To find the exact value of [tex]\(\tan \left(-\frac{\pi}{12}\right)\)[/tex], let’s use a trigonometric identity and the known values of tangent for specific angles.
Step 1: Recognize that [tex]\(-\frac{\pi}{12}\)[/tex] can be represented as the difference of two angles for which the tangent values are known. Specifically:
[tex]\[
-\frac{\pi}{12} = \frac{\pi}{6} - \frac{\pi}{4}
\][/tex]
Step 2: Apply the tangent difference identity, which states that:
[tex]\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
\][/tex]
Here, [tex]\(A = \frac{\pi}{6}\)[/tex] and [tex]\(B = \frac{\pi}{4}\)[/tex].
Step 3: Substitute the known tangent values for [tex]\(\frac{\pi}{6}\)[/tex] and [tex]\(\frac{\pi}{4}\)[/tex]:
[tex]\[
\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\quad \text{and} \quad \tan\left(\frac{\pi}{4}\right) = 1
\][/tex]
Step 4: Substitute these values into the tangent difference identity:
[tex]\[
\tan\left(\frac{\pi}{6} - \frac{\pi}{4}\right) = \frac{\frac{1}{\sqrt{3}} - 1}{1 + \frac{1}{\sqrt{3}} \cdot 1}
\][/tex]
Step 5: Simplify the expression:
[tex]\[
\tan\left(-\frac{\pi}{12}\right) = \frac{\frac{1}{\sqrt{3}} - 1}{1 + \frac{1}{\sqrt{3}}}
\][/tex]
Step 6: Simplify the numerator and the denominator separately:
[tex]\[
\text{Numerator: } \frac{1}{\sqrt{3}} - 1 = \frac{1 - \sqrt{3}}{\sqrt{3}}
\][/tex]
[tex]\[
\text{Denominator: } 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}
\][/tex]
Step 7: Combine these simplified forms into the fraction:
[tex]\[
\tan\left(-\frac{\pi}{12}\right) = \frac{\frac{1 - \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{1 - \sqrt{3}}{\sqrt{3} + 1}
\][/tex]
Step 8: Rationalize the denominator:
Multiply the numerator and the denominator by the conjugate of the denominator [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[
\tan\left(-\frac{\pi}{12}\right) = \frac{(1 - \sqrt{3})(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}
\][/tex]
[tex]\[
\text{Denominator: } (\sqrt{3} + 1)(\sqrt{3} - 1) = 3 - 1 = 2
\][/tex]
[tex]\[
\text{Numerator: } (1 - \sqrt{3})(\sqrt{3} - 1) = 1 \cdot \sqrt{3} - 1 \cdot 1 - \sqrt{3} \cdot \sqrt{3} + \sqrt{3} \cdot 1 = \sqrt{3} - 1 - 3 + \sqrt{3} = -2 + 2\sqrt{3}
\][/tex]
Thus, we have:
[tex]\[
\tan\left(-\frac{\pi}{12}\right) = \frac{-2 + 2\sqrt{3}}{2}
\][/tex]
Step 9: Simplify by cancelling out the common factor of 2 in the numerator:
[tex]\[
\tan\left(-\frac{\pi}{12}\right) = -1 + \sqrt{3}
\][/tex]
Therefore, the exact value of [tex]\(\tan \left(-\frac{\pi}{12}\right)\)[/tex] is:
[tex]\[
\boxed{-2 + \sqrt{3}}
\][/tex]
