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