What is the value of `tan^(-1)((m)/(n))-tan^(-1)((m-n)/(m+n))?`

Options

A. `pi`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `(pi)/(3)`

Answer

To solve the problem \( \tan^{-1}\left(\frac{m}{n}\right) - \tan^{-1}\left(\frac{m-n}{m+n}\right) \), we can use the properties of inverse trigonometric functions, specifically the formula for the difference of two arctangents.

### Step-by-Step Solution:

1. **Let \( x = \tan^{-1}\left(\frac{m}{n}\right) \)**:
This means that \( \tan(x) = \frac{m}{n} \).

2. **Use the formula for the difference of arctangents**:
The formula states:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
Here, we will set \( a = \frac{m}{n} \) and \( b = \frac{m-n}{m+n} \).

3. **Calculate \( a - b \)**:
\[ a - b = \frac{m}{n} - \frac{m-n}{m+n} \]
To combine these fractions, we find a common denominator:
\[ = \frac{m(m+n) - n(m-n)}{n(m+n)} \]
Simplifying the numerator:
\[ = \frac{m^2 + mn - nm + n^2}{n(m+n)} = \frac{m^2 + n^2}{n(m+n)} \]

4. **Calculate \( 1 + ab \)**:
\[ ab = \left(\frac{m}{n}\right)\left(\frac{m-n}{m+n}\right) = \frac{m(m-n)}{n(m+n)} \]
Therefore,
\[ 1 + ab = 1 + \frac{m(m-n)}{n(m+n)} = \frac{n(m+n) + m(m-n)}{n(m+n)} = \frac{nm + n^2 + m^2 - mn}{n(m+n)} = \frac{m^2 + n^2}{n(m+n)} \]

5. **Combine the results**:
Now we can substitute back into the arctangent difference formula:
\[ \tan^{-1}\left(\frac{a - b}{1 + ab}\right) = \tan^{-1}\left(\frac{\frac{m^2 + n^2}{n(m+n)}}{\frac{m^2 + n^2}{n(m+n)}}\right) = \tan^{-1}(1) \]

6. **Final result**:
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we conclude:
\[ \tan^{-1}\left(\frac{m}{n}\right) - \tan^{-1}\left(\frac{m-n}{m+n}\right) = \frac{\pi}{4} \]

### Conclusion:
The value of \( \tan^{-1}\left(\frac{m}{n}\right) - \tan^{-1}\left(\frac{m-n}{m+n}\right) \) is \( \frac{\pi}{4} \).