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To determine which option represents the exact value of [tex]\(\tan \left(\frac{\pi}{12}\right)\)[/tex], we start by understanding the tangent value for the angle [tex]\(\frac{\pi}{12}\)[/tex].
The angle [tex]\(\frac{\pi}{12}\)[/tex] radians can be expressed in degrees as [tex]\(15^\circ\)[/tex] since [tex]\(\frac{180^\circ}{12} = 15^\circ\)[/tex].
Using known trigonometric identities, one approach to find [tex]\(\tan 15^\circ\)[/tex] is recognizing that [tex]\(15^\circ\)[/tex] can be represented as the difference between two familiar angles: [tex]\(45^\circ\)[/tex] and [tex]\(30^\circ\)[/tex]. Therefore, we can use the tangent subtraction formula:
[tex]\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \][/tex]
For [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 30^\circ\)[/tex]:
[tex]\[ \tan 15^\circ = \tan (45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \][/tex]
We know the tangent values:
[tex]\[ \tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \][/tex]
Substituting these into the formula gives us:
[tex]\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \][/tex]
To simplify the fraction, we'll first combine the terms in the numerator and denominator to have a common denominator of [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \][/tex]
Next, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} \][/tex]
[tex]\[ \tan 15^\circ = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Thus, we have:
[tex]\[ \tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3} \][/tex]
We now compare this result with the given options:
1. [tex]\(\frac{1}{2 + \sqrt{3}}\)[/tex]
2. [tex]\(\frac{1}{\sqrt{3}}\)[/tex]
3. [tex]\(\frac{3}{\sqrt{3}}\)[/tex]
4. [tex]\(\frac{1}{2 - \sqrt{3}}\)[/tex]
Evaluating the first option:
[tex]\[ \frac{1}{2 + \sqrt{3}} \][/tex]
We need to rationalize the denominator by multiplying by [tex]\(2 - \sqrt{3}\)[/tex]:
[tex]\[ \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \][/tex]
Now, we see that:
[tex]\[ \tan \left(\frac{\pi}{12}\right) = 2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}} \][/tex]
Thus, the correct answer is the first option:
[tex]\[ \boxed{\frac{1}{2 + \sqrt{3}}} \][/tex]
The angle [tex]\(\frac{\pi}{12}\)[/tex] radians can be expressed in degrees as [tex]\(15^\circ\)[/tex] since [tex]\(\frac{180^\circ}{12} = 15^\circ\)[/tex].
Using known trigonometric identities, one approach to find [tex]\(\tan 15^\circ\)[/tex] is recognizing that [tex]\(15^\circ\)[/tex] can be represented as the difference between two familiar angles: [tex]\(45^\circ\)[/tex] and [tex]\(30^\circ\)[/tex]. Therefore, we can use the tangent subtraction formula:
[tex]\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \][/tex]
For [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 30^\circ\)[/tex]:
[tex]\[ \tan 15^\circ = \tan (45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \][/tex]
We know the tangent values:
[tex]\[ \tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \][/tex]
Substituting these into the formula gives us:
[tex]\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \][/tex]
To simplify the fraction, we'll first combine the terms in the numerator and denominator to have a common denominator of [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \][/tex]
Next, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} \][/tex]
[tex]\[ \tan 15^\circ = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Thus, we have:
[tex]\[ \tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3} \][/tex]
We now compare this result with the given options:
1. [tex]\(\frac{1}{2 + \sqrt{3}}\)[/tex]
2. [tex]\(\frac{1}{\sqrt{3}}\)[/tex]
3. [tex]\(\frac{3}{\sqrt{3}}\)[/tex]
4. [tex]\(\frac{1}{2 - \sqrt{3}}\)[/tex]
Evaluating the first option:
[tex]\[ \frac{1}{2 + \sqrt{3}} \][/tex]
We need to rationalize the denominator by multiplying by [tex]\(2 - \sqrt{3}\)[/tex]:
[tex]\[ \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \][/tex]
Now, we see that:
[tex]\[ \tan \left(\frac{\pi}{12}\right) = 2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}} \][/tex]
Thus, the correct answer is the first option:
[tex]\[ \boxed{\frac{1}{2 + \sqrt{3}}} \][/tex]
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