To find the exact value of [tex]\(\tan \frac{5 \pi}{12}\)[/tex], we can use the half-angle identities. Recall that the half-angle identity for tangent is given by:
[tex]\[
\tan \frac{\theta}{2} = \frac{1 - \cos{\theta}}{\sin{\theta}}
\][/tex]
or alternatively,
[tex]\[
\tan \frac{\theta}{2} = \frac{\sin{\theta}}{1 + \cos{\theta}}
\][/tex]
Firstly, let's begin by rewriting [tex]\(\frac{5 \pi}{12}\)[/tex] in terms of a known angle using the sum of angles identity. Notice that:
[tex]\[
\frac{5 \pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}
\][/tex]
Now we use the tangent addition formula:
[tex]\[
\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
\][/tex]
Let [tex]\( A = \frac{\pi}{4} \)[/tex] and [tex]\( B = \frac{\pi}{6} \)[/tex].
We know:
[tex]\[
\tan \frac{\pi}{4} = 1 \quad \text{and} \quad \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}
\][/tex]
Substitute these into the tangent addition formula:
[tex]\[
\tan \left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \frac{\tan \frac{\pi}{4} + \tan \frac{\pi}{6}}{1 - \tan \frac{\pi}{4} \tan \frac{\pi}{6}}
= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}}
= \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}
= \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}
= \frac{\sqrt{3} + 1}{\sqrt{3} - 1}
\][/tex]
We further simplify this by rationalizing the denominator:
[tex]\[
\frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)}
= \frac{3 + 2\sqrt{3} + 1}{3 - 1}
= \frac{4 + 2\sqrt{3}}{2}
= 2 + \sqrt{3}
\][/tex]
Therefore, the exact value of [tex]\(\tan \frac{5 \pi}{12}\)[/tex] is:
[tex]\[ C) 2 + \sqrt{3} \][/tex]
