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To solve the equation [tex]\(\frac{\cot \theta - 1}{\cot \theta + 1} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}}\)[/tex], we need to manipulate the trigonometric expressions and solve for [tex]\(\theta\)[/tex]. Here's a step-by-step solution:
### Step 1: Simplify the right-hand side fraction
First, simplify the fraction on the right-hand side by rationalizing the denominator:
[tex]\[ \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{(1 - \sqrt{3})^2}{(1 + \sqrt{3})(1 - \sqrt{3})} \][/tex]
The denominator simplifies to:
[tex]\[ (1 + \sqrt{3})(1 - \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2 \][/tex]
The numerator simplifies to:
[tex]\[ (1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + (\sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3} \][/tex]
Therefore, the right-hand side simplifies to:
[tex]\[ \frac{4 - 2\sqrt{3}}{-2} = \frac{4}{-2} - \frac{2\sqrt{3}}{-2} = -2 + \sqrt{3} \][/tex]
So the equation becomes:
[tex]\[ \frac{\cot \theta - 1}{\cot \theta + 1} = -2 + \sqrt{3} \][/tex]
### Step 2: Let [tex]\( \cot \theta = x \)[/tex]
Let [tex]\( x = \cot \theta \)[/tex], the equation becomes:
[tex]\[ \frac{x - 1}{x + 1} = -2 + \sqrt{3} \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
We can set up the equation and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 1 = (-2 + \sqrt{3})(x + 1) \][/tex]
Expand the right-hand side:
[tex]\[ x - 1 = -2x - 2 + \sqrt{3}x + \sqrt{3} \][/tex]
Bring all terms involving [tex]\( x \)[/tex] to one side and constant terms to the other:
[tex]\[ x + 2x - \sqrt{3}x = -2 + \sqrt{3} - 1 \][/tex]
Combine like terms:
[tex]\[ (1 + 2 - \sqrt{3})x = \sqrt{3} - 3 \][/tex]
Thus, we have:
[tex]\[ (3 - \sqrt{3})x = \sqrt{3} - 3 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\sqrt{3} - 3}{3 - \sqrt{3}} \][/tex]
### Step 4: Rationalize the denominator
To simplify [tex]\( x \)[/tex], rationalize the denominator:
[tex]\[ x = \frac{\sqrt{3} - 3}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{(\sqrt{3} - 3)(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} \][/tex]
The denominator simplifies to:
[tex]\[ (3 - \sqrt{3})(3 + \sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6 \][/tex]
The numerator simplifies to:
[tex]\[ (\sqrt{3} - 3)(3 + \sqrt{3}) = 3\sqrt{3} + 3 - 9 - 3\sqrt{3} = 3 - 9 = -6 \][/tex]
Thus:
[tex]\[ x = \frac{-6}{6} = -1 \][/tex]
### Step 5: Find [tex]\(\theta\)[/tex]
Since [tex]\( x = \cot \theta \)[/tex] and [tex]\( x = -1 \)[/tex]:
[tex]\[ \cot \theta = -1 \][/tex]
This relationship holds true for:
[tex]\[ \theta = 135^\circ \text{ or } \frac{3\pi}{4} \text{ in standard position} \][/tex]
Nevertheless, we need to find the acute angle. Recall that [tex]\(\cot(180^\circ - \theta) = -\cot(\theta)\)[/tex], and thus:
[tex]\[ \theta = 45^\circ \text{ or } \frac{\pi}{4} \][/tex]
Therefore, the acute angle [tex]\(\theta\)[/tex] is:
[tex]\[ \theta = 45^\circ \text{ or } \frac{\pi}{4} \][/tex]
This completes the detailed, step-by-step solution.
### Step 1: Simplify the right-hand side fraction
First, simplify the fraction on the right-hand side by rationalizing the denominator:
[tex]\[ \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{(1 - \sqrt{3})^2}{(1 + \sqrt{3})(1 - \sqrt{3})} \][/tex]
The denominator simplifies to:
[tex]\[ (1 + \sqrt{3})(1 - \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2 \][/tex]
The numerator simplifies to:
[tex]\[ (1 - \sqrt{3})^2 = 1 - 2\sqrt{3} + (\sqrt{3})^2 = 1 - 2\sqrt{3} + 3 = 4 - 2\sqrt{3} \][/tex]
Therefore, the right-hand side simplifies to:
[tex]\[ \frac{4 - 2\sqrt{3}}{-2} = \frac{4}{-2} - \frac{2\sqrt{3}}{-2} = -2 + \sqrt{3} \][/tex]
So the equation becomes:
[tex]\[ \frac{\cot \theta - 1}{\cot \theta + 1} = -2 + \sqrt{3} \][/tex]
### Step 2: Let [tex]\( \cot \theta = x \)[/tex]
Let [tex]\( x = \cot \theta \)[/tex], the equation becomes:
[tex]\[ \frac{x - 1}{x + 1} = -2 + \sqrt{3} \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex]
We can set up the equation and solve for [tex]\( x \)[/tex]:
[tex]\[ x - 1 = (-2 + \sqrt{3})(x + 1) \][/tex]
Expand the right-hand side:
[tex]\[ x - 1 = -2x - 2 + \sqrt{3}x + \sqrt{3} \][/tex]
Bring all terms involving [tex]\( x \)[/tex] to one side and constant terms to the other:
[tex]\[ x + 2x - \sqrt{3}x = -2 + \sqrt{3} - 1 \][/tex]
Combine like terms:
[tex]\[ (1 + 2 - \sqrt{3})x = \sqrt{3} - 3 \][/tex]
Thus, we have:
[tex]\[ (3 - \sqrt{3})x = \sqrt{3} - 3 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\sqrt{3} - 3}{3 - \sqrt{3}} \][/tex]
### Step 4: Rationalize the denominator
To simplify [tex]\( x \)[/tex], rationalize the denominator:
[tex]\[ x = \frac{\sqrt{3} - 3}{3 - \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{(\sqrt{3} - 3)(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} \][/tex]
The denominator simplifies to:
[tex]\[ (3 - \sqrt{3})(3 + \sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6 \][/tex]
The numerator simplifies to:
[tex]\[ (\sqrt{3} - 3)(3 + \sqrt{3}) = 3\sqrt{3} + 3 - 9 - 3\sqrt{3} = 3 - 9 = -6 \][/tex]
Thus:
[tex]\[ x = \frac{-6}{6} = -1 \][/tex]
### Step 5: Find [tex]\(\theta\)[/tex]
Since [tex]\( x = \cot \theta \)[/tex] and [tex]\( x = -1 \)[/tex]:
[tex]\[ \cot \theta = -1 \][/tex]
This relationship holds true for:
[tex]\[ \theta = 135^\circ \text{ or } \frac{3\pi}{4} \text{ in standard position} \][/tex]
Nevertheless, we need to find the acute angle. Recall that [tex]\(\cot(180^\circ - \theta) = -\cot(\theta)\)[/tex], and thus:
[tex]\[ \theta = 45^\circ \text{ or } \frac{\pi}{4} \][/tex]
Therefore, the acute angle [tex]\(\theta\)[/tex] is:
[tex]\[ \theta = 45^\circ \text{ or } \frac{\pi}{4} \][/tex]
This completes the detailed, step-by-step solution.
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