`tan^(-1)((sqrt(2)+1)/(sqrt(2)-1)) - tan^(-1)(sqrt(2)/2)`=

Options

A. `pi/3`
B. `pi/6`
C. `pi/4`
D. `pi/5`

Answer

To solve the problem \( \tan^{-1}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) - \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) \), we will use the formula for the difference of two inverse tangents:

\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1\left(\frac{A - B}{1 + AB}\right) \]

### Step 1: Identify A and B
Let:
- \( A = \frac{\sqrt{2}+1}{\sqrt{2}-1} \)
- \( B = \frac{\sqrt{2}}{2} \)

### Step 2: Calculate A - B
Now we need to compute \( A - B \):

\[ A - B = \frac{\sqrt{2}+1}{\sqrt{2}-1} - \frac{\sqrt{2}}{2} \]

To subtract these fractions, we need a common denominator. The common denominator will be \( 2(\sqrt{2}-1) \):

\[ A - B = \frac{2(\sqrt{2}+1) - \sqrt{2}(\sqrt{2}-1)}{2(\sqrt{2}-1)} \]

### Step 3: Simplify the numerator
Now, simplify the numerator:

\[ 2(\sqrt{2}+1) = 2\sqrt{2} + 2 \]
\[ \sqrt{2}(\sqrt{2}-1) = 2 - \sqrt{2} \]

So,

\[ A - B = \frac{(2\sqrt{2} + 2) - (2 - \sqrt{2})}{2(\sqrt{2}-1)} = \frac{2\sqrt{2} + 2 - 2 + \sqrt{2}}{2(\sqrt{2}-1)} = \frac{3\sqrt{2}}{2(\sqrt{2}-1)} \]

### Step 4: Calculate 1 + AB
Next, we need to calculate \( 1 + AB \):

\[ AB = \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\left(\frac{\sqrt{2}}{2}\right) = \frac{(\sqrt{2}+1)\sqrt{2}}{2(\sqrt{2}-1)} = \frac{2 + \sqrt{2}}{2(\sqrt{2}-1)} \]

Now, add 1:

\[ 1 + AB = 1 + \frac{2 + \sqrt{2}}{2(\sqrt{2}-1)} = \frac{2(\sqrt{2}-1) + 2 + \sqrt{2}}{2(\sqrt{2}-1)} = \frac{2\sqrt{2} - 2 + 2 + \sqrt{2}}{2(\sqrt{2}-1)} = \frac{3\sqrt{2}}{2(\sqrt{2}-1)} \]

### Step 5: Substitute into the formula
Now we can substitute back into the formula:

\[ \tan^{-1}\left(\frac{A - B}{1 + AB}\right) = \tan^{-1}\left(\frac{\frac{3\sqrt{2}}{2(\sqrt{2}-1)}}{\frac{3\sqrt{2}}{2(\sqrt{2}-1)}}\right) = \tan^{-1}(1) \]

### Step 6: Final result
Since \( \tan^{-1}(1) = \frac{\pi}{4} \):

\[ \tan^{-1}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) - \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \]