`tan^(-1)5+tan^(-1) 3-cot^(-1) "4/7=`

Options

A. `pi//4`
B. `pi//2`
C. `3pi//4`
D. none

Answer

To solve the equation \( \tan^{-1}(5) + \tan^{-1}(3) - \cot^{-1}\left(\frac{4}{7}\right) \), we will use the properties of inverse trigonometric functions.

### Step-by-Step Solution:

1. **Use the formula for the sum of inverse tangents**:
We know that:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \quad \text{if } xy < 1 \]
Here, \( x = 5 \) and \( y = 3 \).

2. **Calculate \( xy \)**:
\[ xy = 5 \times 3 = 15 \]
Since \( 15 > 1 \), we will use the alternative formula:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) + \pi \]

3. **Calculate \( \tan^{-1}(5) + \tan^{-1}(3) \)**:
\[ \tan^{-1}(5) + \tan^{-1}(3) = \tan^{-1}\left(\frac{5 + 3}{1 - 15}\right) + \pi = \tan^{-1}\left(\frac{8}{-14}\right) + \pi = \tan^{-1}\left(-\frac{4}{7}\right) + \pi \]

4. **Combine with \( -\cot^{-1}\left(\frac{4}{7}\right) \)**:
We know that:
\[ \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x) \]
Therefore:
\[ -\cot^{-1}\left(\frac{4}{7}\right) = -\left(\frac{\pi}{2} - \tan^{-1}\left(\frac{4}{7}\right)\right) = -\frac{\pi}{2} + \tan^{-1}\left(\frac{4}{7}\right) \]

5. **Combine all parts**:
\[ \tan^{-1}\left(-\frac{4}{7}\right) + \pi - \frac{\pi}{2} + \tan^{-1}\left(\frac{4}{7}\right) \]
Since \( \tan^{-1}\left(-\frac{4}{7}\right) + \tan^{-1}\left(\frac{4}{7}\right) = 0 \):
\[ \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

### Final Answer:
\[ \frac{\pi}{2} \]