`cos^(-1) (cos (2 cot^(-1) (sqrt2 -1)))` is equal to

Options

A. `sqrt2 -1`
B. `(pi)/(4)`
C. `(3pi)/(4)`
D. none of these

Explanation

To solve the expression \( \cos^{-1}(\cos(2 \cot^{-1}(\sqrt{2} - 1))) \), we will follow these steps:

### Step 1: Simplify \( \cot^{-1}(\sqrt{2} - 1) \)

We start by evaluating \( \cot^{-1}(\sqrt{2} - 1) \). We know that \( \cot^{-1}(x) \) gives us an angle whose cotangent is \( x \).

Let \( \theta = \cot^{-1}(\sqrt{2} - 1) \). This means:
\[ \cot(\theta) = \sqrt{2} - 1 \]

### Step 2: Find \( \tan(\theta) \)

Using the identity \( \tan(\theta) = \frac{1}{\cot(\theta)} \):
\[ \tan(\theta) = \frac{1}{\sqrt{2} - 1} \]

To simplify this, we can rationalize the denominator:
\[ \tan(\theta) = \frac{1(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \]

### Step 3: Find \( 2 \cot^{-1}(\sqrt{2} - 1) \)

Now we need to find \( 2\theta \):
\[ 2\theta = 2 \cot^{-1}(\sqrt{2} - 1) \]

Using the double angle formula for cotangent:
\[ \cot(2\theta) = \frac{\cot^2(\theta) - 1}{2\cot(\theta)} \]

Substituting \( \cot(\theta) = \sqrt{2} - 1 \):
\[ \cot(2\theta) = \frac{(\sqrt{2} - 1)^2 - 1}{2(\sqrt{2} - 1)} \]
Calculating \( (\sqrt{2} - 1)^2 \):
\[ (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \]
Thus,
\[ \cot(2\theta) = \frac{(3 - 2\sqrt{2}) - 1}{2(\sqrt{2} - 1)} = \frac{2 - 2\sqrt{2}}{2(\sqrt{2} - 1)} = \frac{1 - \sqrt{2}}{\sqrt{2} - 1} = -1 \]

### Step 4: Find \( 2\theta \)

Since \( \cot(2\theta) = -1 \), we know:
\[ 2\theta = \frac{3\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 5: Find \( \cos^{-1}(\cos(2\theta)) \)

Now we substitute back into our original expression:
\[ \cos^{-1}(\cos(2\theta)) = \cos^{-1}(\cos(\frac{3\pi}{4})) \]

Since \( \cos^{-1}(\cos(x)) = x \) if \( x \) is in the range \( [0, \pi] \), we have:
\[ \cos^{-1}(\cos(\frac{3\pi}{4})) = \frac{3\pi}{4} \]

### Final Answer

Thus, the final answer is:
\[ \cos^{-1}(\cos(2 \cot^{-1}(\sqrt{2} - 1))) = \frac{3\pi}{4} \]