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To find the exact value of [tex]\(\tan \left(\frac{7 \pi}{12}\right)\)[/tex], we can use the angle addition formula for tangent. Let's start by recognizing that [tex]\(\frac{7 \pi}{12}\)[/tex] radians is equivalent to [tex]\(105^\circ\)[/tex] in degrees.
We can express [tex]\(105^\circ\)[/tex] as the sum of two angles that are familiar:
[tex]\[ 105^\circ = 45^\circ + 60^\circ \][/tex]
We can use the tangent addition formula:
[tex]\[ \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a) \tan(b)} \][/tex]
Here, let [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 60^\circ\)[/tex].
First, we find the tangents of these angles:
[tex]\[ \tan(45^\circ) = 1 \][/tex]
[tex]\[ \tan(60^\circ) = \sqrt{3} \][/tex]
Now, apply the tangent addition formula with [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 60^\circ\)[/tex]:
[tex]\[ \tan(105^\circ) = \frac{\tan(45^\circ) + \tan(60^\circ)}{1 - \tan(45^\circ) \tan(60^\circ)} = \frac{1 + \sqrt{3}}{1 - 1 \cdot \sqrt{3}} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \][/tex]
As required, we need to rationalize the denominator:
[tex]\[ \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \][/tex]
Multiplying both the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{(1 + \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \][/tex]
Simplify the denominator using the difference of squares [tex]\( (a - b)(a + b) = a^2 - b^2 \)[/tex]:
[tex]\[ = \frac{(1 + \sqrt{3})^2}{1 - (\sqrt{3})^2} = \frac{1 + 2\sqrt{3} + 3}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} \][/tex]
Simplify the fraction:
[tex]\[ = \frac{4 + 2\sqrt{3}}{-2} = \frac{4}{-2} + \frac{2\sqrt{3}}{-2} = -2 - \sqrt{3} \][/tex]
Thus, the exact value of [tex]\(\tan \left(\frac{7\pi}{12}\right)\)[/tex] after rationalizing the denominator is:
[tex]\[ -2 - \sqrt{3} \][/tex]
To ensure our computations are correct, we recognize the formula results and computations match our numerical value:
[tex]\[ (-3.7320508075688803, -1.9999999999999976, 0.5358983848622443) \][/tex]
Where our rationalized denominator and numerator give:
[tex]\[ = -3.7320508075688803 \][/tex]
Hence:
[tex]\[ \tan \left(\frac{7 \pi}{12}\right) = -3.7320508075688803 \][/tex]
Which confirms our precise steps in evaluating [tex]\(\tan \left(\frac{7 \pi}{12}\right)\)[/tex]!
We can express [tex]\(105^\circ\)[/tex] as the sum of two angles that are familiar:
[tex]\[ 105^\circ = 45^\circ + 60^\circ \][/tex]
We can use the tangent addition formula:
[tex]\[ \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a) \tan(b)} \][/tex]
Here, let [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 60^\circ\)[/tex].
First, we find the tangents of these angles:
[tex]\[ \tan(45^\circ) = 1 \][/tex]
[tex]\[ \tan(60^\circ) = \sqrt{3} \][/tex]
Now, apply the tangent addition formula with [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 60^\circ\)[/tex]:
[tex]\[ \tan(105^\circ) = \frac{\tan(45^\circ) + \tan(60^\circ)}{1 - \tan(45^\circ) \tan(60^\circ)} = \frac{1 + \sqrt{3}}{1 - 1 \cdot \sqrt{3}} = \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \][/tex]
As required, we need to rationalize the denominator:
[tex]\[ \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \][/tex]
Multiplying both the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{(1 + \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \][/tex]
Simplify the denominator using the difference of squares [tex]\( (a - b)(a + b) = a^2 - b^2 \)[/tex]:
[tex]\[ = \frac{(1 + \sqrt{3})^2}{1 - (\sqrt{3})^2} = \frac{1 + 2\sqrt{3} + 3}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} \][/tex]
Simplify the fraction:
[tex]\[ = \frac{4 + 2\sqrt{3}}{-2} = \frac{4}{-2} + \frac{2\sqrt{3}}{-2} = -2 - \sqrt{3} \][/tex]
Thus, the exact value of [tex]\(\tan \left(\frac{7\pi}{12}\right)\)[/tex] after rationalizing the denominator is:
[tex]\[ -2 - \sqrt{3} \][/tex]
To ensure our computations are correct, we recognize the formula results and computations match our numerical value:
[tex]\[ (-3.7320508075688803, -1.9999999999999976, 0.5358983848622443) \][/tex]
Where our rationalized denominator and numerator give:
[tex]\[ = -3.7320508075688803 \][/tex]
Hence:
[tex]\[ \tan \left(\frac{7 \pi}{12}\right) = -3.7320508075688803 \][/tex]
Which confirms our precise steps in evaluating [tex]\(\tan \left(\frac{7 \pi}{12}\right)\)[/tex]!
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