`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

Answer

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

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

We start by letting \( x = \cot^{-1}(\sqrt{2} - 1) \). Therefore, we have:
\[ \cot(x) = \sqrt{2} - 1 \]
To find \( \tan(x) \), we use the identity \( \tan(x) = \frac{1}{\cot(x)} \):
\[ \tan(x) = \frac{1}{\sqrt{2} - 1} \]
Next, we rationalize the denominator:
\[ \tan(x) = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \]

### Step 2: Use the double angle formula for tangent

Now, we can find \( 2 \cot^{-1}(\sqrt{2} - 1) \):
\[ 2 \cot^{-1}(\sqrt{2} - 1) = 2 \tan^{-1}(\sqrt{2} + 1) \]
Using the double angle formula for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Let \( \theta = \tan^{-1}(\sqrt{2} + 1) \):
\[ \tan(2\theta) = \frac{2(\sqrt{2} + 1)}{1 - (\sqrt{2} + 1)^2} \]
Calculating \( (\sqrt{2} + 1)^2 \):
\[ (\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2} \]
Thus,
\[ 1 - (\sqrt{2} + 1)^2 = 1 - (3 + 2\sqrt{2}) = -2 - 2\sqrt{2} \]
Now substituting back:
\[ \tan(2\theta) = \frac{2(\sqrt{2} + 1)}{-2 - 2\sqrt{2}} = -\frac{\sqrt{2} + 1}{1 + \sqrt{2}} = -1 \]

### Step 3: Find the angle corresponding to \( \tan(2\theta) = -1 \)

The angle whose tangent is -1 is:
\[ 2\theta = \frac{3\pi}{4} \quad \text{(since it is in the second quadrant)} \]

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

Thus,
\[ \theta = \frac{3\pi}{8} \]

### Step 5: Substitute back into the original expression

Now we substitute back to find:
\[ \cos^{-1}(\cos(2 \cot^{-1}(\sqrt{2} - 1))) = \cos^{-1}(\cos(\frac{3\pi}{4})) \]
Since \( \cos^{-1}(\cos(x)) = x \) for \( x \) 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:
\[ \frac{3\pi}{4} \]