Solve `: tan ((pi)/(4)+ theta) + tan ((pi)/(4) - theta) = 4`

Answer

To solve the equation \( \tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = 4 \), we can follow these steps:

### Step 1: Apply the tangent addition and subtraction formulas
Using the tangent addition and subtraction formulas, we have:
\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan\frac{\pi}{4} + \tan\theta}{1 - \tan\frac{\pi}{4} \tan\theta} \]
\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{\tan\frac{\pi}{4} - \tan\theta}{1 + \tan\frac{\pi}{4} \tan\theta} \]
Since \( \tan\frac{\pi}{4} = 1 \), we can substitute this into the formulas:
\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]
\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta} \]

### Step 2: Substitute into the original equation
Substituting these results into the original equation gives:
\[ \frac{1 + \tan\theta}{1 - \tan\theta} + \frac{1 - \tan\theta}{1 + \tan\theta} = 4 \]

### Step 3: Find a common denominator
The common denominator for the left-hand side is \( (1 - \tan\theta)(1 + \tan\theta) \):
\[ \frac{(1 + \tan\theta)^2 + (1 - \tan\theta)^2}{(1 - \tan\theta)(1 + \tan\theta)} = 4 \]

### Step 4: Expand the numerator
Expanding the numerator:
\[ (1 + \tan\theta)^2 = 1 + 2\tan\theta + \tan^2\theta \]
\[ (1 - \tan\theta)^2 = 1 - 2\tan\theta + \tan^2\theta \]
Adding these gives:
\[ 1 + 2\tan\theta + \tan^2\theta + 1 - 2\tan\theta + \tan^2\theta = 2 + 2\tan^2\theta \]
Thus, we have:
\[ \frac{2 + 2\tan^2\theta}{1 - \tan^2\theta} = 4 \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ 2 + 2\tan^2\theta = 4(1 - \tan^2\theta) \]
Expanding the right-hand side:
\[ 2 + 2\tan^2\theta = 4 - 4\tan^2\theta \]

### Step 6: Rearrange the equation
Bringing all terms involving \( \tan^2\theta \) to one side:
\[ 2\tan^2\theta + 4\tan^2\theta = 4 - 2 \]
\[ 6\tan^2\theta = 2 \]
\[ \tan^2\theta = \frac{1}{3} \]

### Step 7: Solve for \( \tan\theta \)
Taking the square root gives:
\[ \tan\theta = \frac{1}{\sqrt{3}} \quad \text{or} \quad \tan\theta = -\frac{1}{\sqrt{3}} \]

### Step 8: Find the general solutions
The angles corresponding to \( \tan\theta = \frac{1}{\sqrt{3}} \) are:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]
The angles corresponding to \( \tan\theta = -\frac{1}{\sqrt{3}} \) are:
\[ \theta = \frac{5\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

### Final Answer
Thus, the general solutions are:
\[ \theta = \frac{\pi}{6} + n\pi \quad \text{and} \quad \theta = \frac{5\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]