`tan^(-1)n ,tan^(-1)(n+1)` and `tan^(-1)(n+2),n in N ,` are the angles of a triangle if `n=` 1 (b) 2 (c) 3 (d) None of these

Options

A. `-1`
B. 1
C. 2
D. None of these

Answer

To determine the value of \( n \) such that \( \tan^{-1}(n) \), \( \tan^{-1}(n+1) \), and \( \tan^{-1}(n+2) \) are the angles of a triangle, we need to use the property that the sum of the angles in a triangle is \( 180^\circ \) or \( \pi \) radians.

### Step-by-step Solution:

1. **Set up the equation**:
We know that the sum of the angles must equal \( \pi \):
\[ \tan^{-1}(n) + \tan^{-1}(n+1) + \tan^{-1}(n+2) = \pi \]

2. **Rearrange the equation**:
We can rearrange this to isolate \( \tan^{-1}(n) \):
\[ \tan^{-1}(n+1) + \tan^{-1}(n+2) = \pi - \tan^{-1}(n) \]

3. **Use the tangent addition formula**:
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \quad \text{if } ab < 1 \]
Applying this to our equation:
\[ \tan^{-1}(n+1) + \tan^{-1}(n+2) = \tan^{-1}\left(\frac{(n+1)+(n+2)}{1-(n+1)(n+2)}\right) \]
This simplifies to:
\[ \tan^{-1}\left(\frac{2n + 3}{1 - (n^2 + 3n + 2)}\right) \]

4. **Set the two expressions equal**:
Now we have:
\[ \tan^{-1}\left(\frac{2n + 3}{1 - (n^2 + 3n + 2)}\right) = \tan^{-1}(-n) \]
This implies:
\[ \frac{2n + 3}{1 - (n^2 + 3n + 2)} = -n \]

5. **Cross-multiply to eliminate the tangent**:
Cross-multiplying gives:
\[ 2n + 3 = -n(1 - (n^2 + 3n + 2)) \]
Expanding the right side:
\[ 2n + 3 = -n + n^3 + 3n^2 + 2n \]

6. **Rearranging the equation**:
Combine like terms:
\[ 0 = n^3 + 3n^2 + 5n + 3 \]

7. **Finding the roots**:
We need to solve the cubic equation \( n^3 + 3n^2 + 5n + 3 = 0 \). By trying \( n = -1 \):
\[ (-1)^3 + 3(-1)^2 + 5(-1) + 3 = -1 + 3 - 5 + 3 = 0 \]
Thus, \( n = -1 \) is a root.

8. **Conclusion**:
Since \( n \) must be a natural number, and the only real root we found is \( n = -1 \), which is not in \( \mathbb{N} \), we conclude that there are no natural number solutions.

### Final Answer:
The value of \( n \) is **None of these**.