prove that `2tan^-1 1/3+tan^-1 1/7=pi/4`

Options

A. `tan^(-1)(49/29)`
B. `(pi)/(2)`
C. 0
D. `(pi)/(4)`

Answer

To prove that \( 2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Rewrite the expression using the double angle formula
We start with the left-hand side:
\[ 2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) \]
Using the identity \( 2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), we can rewrite \( 2\tan^{-1}\left(\frac{1}{3}\right) \):
\[ 2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}\right) \]

### Step 2: Simplify the expression
Now, calculate the expression inside the arctangent:
\[ = \tan^{-1}\left(\frac{\frac{2}{3}}{1 - \frac{1}{9}}\right) = \tan^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) \]
This simplifies to:
\[ = \tan^{-1}\left(\frac{2}{3} \cdot \frac{9}{8}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 3: Combine with the other arctangent
Now we can rewrite the left-hand side as:
\[ \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{7}\right) \]
Using the identity \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \), we apply this to our expression:
\[ = \tan^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \cdot \frac{1}{7}}\right) \]

### Step 4: Simplify the combined expression
Calculating the numerator:
\[ \frac{3}{4} + \frac{1}{7} = \frac{21 + 4}{28} = \frac{25}{28} \]
Calculating the denominator:
\[ 1 - \frac{3}{4} \cdot \frac{1}{7} = 1 - \frac{3}{28} = \frac{28 - 3}{28} = \frac{25}{28} \]
Thus, we have:
\[ = \tan^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) = \tan^{-1}(1) \]

### Step 5: Final result
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we conclude:
\[ 2\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \frac{\pi}{4} \]
This proves the statement.