Answer:
2 - √3 or tan (15°)
Step-by-step explanation:
Given:
[tex] \tan(\frac{\pi}{12}) [/tex]
To find the exact value of [tex] \sf \tan\left(\frac{\pi}{12}\right) [/tex], we can use the tangent subtraction formula:
[tex] \sf \tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} [/tex]
Let's set [tex] \sf a = \frac{\pi}{4} [/tex] and [tex] \sf b = \frac{\pi}{6} [/tex], so:
[tex] \sf \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi}{12} - \frac{2\pi}{12} = \frac{\pi}{12} [/tex]
Thus, we have:
[tex] \sf \tan\left(\frac{\pi}{12}\right) = \tan\left(\frac{\pi}{4} - \frac{\pi}{6}\right) [/tex]
Using the tangent subtraction formula:
[tex] \sf \tan\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan\left(\frac{\pi}{6}\right)}{1 + \tan\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{6}\right)} [/tex]
We know:
[tex] \sf \tan\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} [/tex]
Substitute these values into the formula:
[tex] \sf \tan\left(\frac{\pi}{12}\right) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} [/tex]
Simplify the numerator and the denominator by multiplying both by [tex] \sf \sqrt{3} [/tex]:
[tex] \sf \tan\left(\frac{\pi}{12}\right) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} [/tex]
To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator [tex] \sf \sqrt{3} - 1 [/tex]:
[tex] \sf \tan\left(\frac{\pi}{12}\right) = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} [/tex]
Simplify further:
[tex] \sf \tan\left(\frac{\pi}{12}\right) = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} [/tex]
So, the exact value of [tex] \sf \tan\left(\frac{\pi}{12}\right) [/tex] is:
[tex] \sf \tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3} [/tex]