Prove that `cot((pi)/(4) + 0) cot ((pi)/(4) - 0) =1`.

Answer

To prove that \( \cot\left(\frac{\pi}{4} + \theta\right) \cot\left(\frac{\pi}{4} - \theta\right) = 1 \), we will start with the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step 1: Write down the LHS
We start with:
\[ \text{LHS} = \cot\left(\frac{\pi}{4} + \theta\right) \cot\left(\frac{\pi}{4} - \theta\right) \]

### Step 2: Use the cotangent addition formula
Recall that:
\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]
For \( A = \frac{\pi}{4} \) and \( B = \theta \), we have:
\[ \cot\left(\frac{\pi}{4} + \theta\right) = \frac{\cot\left(\frac{\pi}{4}\right) \cot(\theta) - 1}{\cot\left(\frac{\pi}{4}\right) + \cot(\theta)} \]
Since \( \cot\left(\frac{\pi}{4}\right) = 1 \), this simplifies to:
\[ \cot\left(\frac{\pi}{4} + \theta\right) = \frac{1 \cdot \cot(\theta) - 1}{1 + \cot(\theta)} = \frac{\cot(\theta) - 1}{1 + \cot(\theta)} \]

Similarly, for \( \cot\left(\frac{\pi}{4} - \theta\right) \):
\[ \cot\left(\frac{\pi}{4} - \theta\right) = \frac{\cot\left(\frac{\pi}{4}\right) \cot(-\theta) - 1}{\cot\left(\frac{\pi}{4}\right) + \cot(-\theta)} \]
Since \( \cot(-\theta) = -\cot(\theta) \), we have:
\[ \cot\left(\frac{\pi}{4} - \theta\right) = \frac{1 \cdot (-\cot(\theta)) - 1}{1 + (-\cot(\theta))} = \frac{-\cot(\theta) - 1}{1 - \cot(\theta)} = \frac{-(\cot(\theta) + 1)}{1 - \cot(\theta)} \]

### Step 3: Substitute back into the LHS
Now substituting both results back into the LHS:
\[ \text{LHS} = \left(\frac{\cot(\theta) - 1}{1 + \cot(\theta)}\right) \left(\frac{-(\cot(\theta) + 1)}{1 - \cot(\theta)}\right) \]

### Step 4: Simplify the expression
Now we simplify:
\[ \text{LHS} = \frac{(\cot(\theta) - 1)(-\cot(\theta) - 1)}{(1 + \cot(\theta))(1 - \cot(\theta))} \]
The numerator becomes:
\[ -(\cot^2(\theta) + \cot(\theta) - \cot(\theta) - 1) = -(\cot^2(\theta) - 1) = 1 - \cot^2(\theta) \]
The denominator becomes:
\[ 1 - \cot^2(\theta) \]
Thus, we have:
\[ \text{LHS} = \frac{1 - \cot^2(\theta)}{1 - \cot^2(\theta)} = 1 \]

### Step 5: Conclusion
Therefore, we have shown that:
\[ \cot\left(\frac{\pi}{4} + \theta\right) \cot\left(\frac{\pi}{4} - \theta\right) = 1 \]
This verifies the identity.