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To solve the expression for [tex]\(\tan \frac{7 \pi}{12}\)[/tex], we can use the trigonometric angle sum identity. Specifically, we can represent [tex]\(\frac{7 \pi}{12}\)[/tex] as a sum of two angles for which the tangent values are known.
1. Step 1: Choosing appropriate angles
We can express [tex]\(\frac{7\pi}{12}\)[/tex] in terms of [tex]\(\frac{\pi}{3}\)[/tex] and [tex]\(\frac{\pi}{4}\)[/tex]:
[tex]\[ \frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4} \][/tex]
2. Step 2: Using the tangent sum identity
The tangent of a sum of two angles [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is given by:
[tex]\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} \][/tex]
Here, [tex]\(A = \frac{\pi}{3}\)[/tex] and [tex]\(B = \frac{\pi}{4}\)[/tex].
3. Step 3: Finding the tangents of [tex]\(\frac{\pi}{3}\)[/tex] and [tex]\(\frac{\pi}{4}\)[/tex]
We know:
[tex]\[ \tan \frac{\pi}{3} = \sqrt{3} \][/tex]
[tex]\[ \tan \frac{\pi}{4} = 1 \][/tex]
4. Step 4: Applying the angle sum identity
Substituting these values into the sum formula:
[tex]\[ \tan \left( \frac{\pi}{3} + \frac{\pi}{4} \right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} \][/tex]
5. Step 5: Simplifying the expression
Simplify the numerator and the denominator:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \][/tex]
6. Step 6: Identifying the underlying expression
The expression [tex]\(\frac{\sqrt{3} + 1}{1 - \sqrt{3}}\)[/tex] can be manipulated to match the form given in the question. Notice that:
[tex]\[ 1 - \sqrt{3} < 0 \][/tex]
So the expression will be negative.
To match the form [tex]\(-\sqrt{\frac{2+\sqrt{?} }{ \square -\sqrt{\square} }}\)[/tex], we rewrite:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = -\sqrt{\frac{2+\sqrt{3}}{1-\sqrt{3}}} \][/tex]
The expression inside the square root should be balanced such that the numerator and denominator fit correctly. Noting that the result was [tex]\(-3.7320508075688776\)[/tex], and matching it to the form given in the problem statement, we can see:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = -\sqrt{\frac{2+\sqrt{3}}{1-\sqrt{3}}} \][/tex]
Hence, we fill the blanks as:
[tex]\[ \tan \frac{7 \pi}{12}=-\sqrt{\frac{2+\sqrt{3}}{1-\sqrt{3}}} \][/tex]
1. Step 1: Choosing appropriate angles
We can express [tex]\(\frac{7\pi}{12}\)[/tex] in terms of [tex]\(\frac{\pi}{3}\)[/tex] and [tex]\(\frac{\pi}{4}\)[/tex]:
[tex]\[ \frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4} \][/tex]
2. Step 2: Using the tangent sum identity
The tangent of a sum of two angles [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is given by:
[tex]\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} \][/tex]
Here, [tex]\(A = \frac{\pi}{3}\)[/tex] and [tex]\(B = \frac{\pi}{4}\)[/tex].
3. Step 3: Finding the tangents of [tex]\(\frac{\pi}{3}\)[/tex] and [tex]\(\frac{\pi}{4}\)[/tex]
We know:
[tex]\[ \tan \frac{\pi}{3} = \sqrt{3} \][/tex]
[tex]\[ \tan \frac{\pi}{4} = 1 \][/tex]
4. Step 4: Applying the angle sum identity
Substituting these values into the sum formula:
[tex]\[ \tan \left( \frac{\pi}{3} + \frac{\pi}{4} \right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} \][/tex]
5. Step 5: Simplifying the expression
Simplify the numerator and the denominator:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \][/tex]
6. Step 6: Identifying the underlying expression
The expression [tex]\(\frac{\sqrt{3} + 1}{1 - \sqrt{3}}\)[/tex] can be manipulated to match the form given in the question. Notice that:
[tex]\[ 1 - \sqrt{3} < 0 \][/tex]
So the expression will be negative.
To match the form [tex]\(-\sqrt{\frac{2+\sqrt{?} }{ \square -\sqrt{\square} }}\)[/tex], we rewrite:
[tex]\[ \tan \left( \frac{7\pi}{12} \right) = -\sqrt{\frac{2+\sqrt{3}}{1-\sqrt{3}}} \][/tex]
The expression inside the square root should be balanced such that the numerator and denominator fit correctly. Noting that the result was [tex]\(-3.7320508075688776\)[/tex], and matching it to the form given in the problem statement, we can see:
[tex]\[ \tan \left( \frac{7 \pi}{12} \right) = -\sqrt{\frac{2+\sqrt{3}}{1-\sqrt{3}}} \][/tex]
Hence, we fill the blanks as:
[tex]\[ \tan \frac{7 \pi}{12}=-\sqrt{\frac{2+\sqrt{3}}{1-\sqrt{3}}} \][/tex]
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