If `a_1, a_2,a_3, ,a_n` is an A.P. with common difference `d ,` then prove that `"tan"[tan^(-1)(d/(1+a_1a_2))+tan^(-1)(d/(1+a_2a_3))+tan^(-1)(d/(11+a_(n-1)a_n))]=((n-1)d)/(1+a_1a_n)`

Explanation

To prove the statement:

\[ \tan\left(\tan^{-1}\left(\frac{d}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{d}{1 + a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{d}{1 + a_{n-1} a_n}\right)\right) = \frac{(n-1)d}{1 + a_1 a_n} \]

where \(a_1, a_2, a_3, \ldots, a_n\) is an arithmetic progression (A.P.) with common difference \(d\).

### Step-by-step Solution:

1. **Understanding the A.P.**:
Since \(a_1, a_2, a_3, \ldots, a_n\) is an A.P. with common difference \(d\):
\[ a_2 = a_1 + d, \quad a_3 = a_1 + 2d, \quad \ldots, \quad a_n = a_1 + (n-1)d \]

2. **Expressing the LHS**:
The left-hand side (LHS) can be expressed as:
\[ \tan\left(\tan^{-1}\left(\frac{d}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{d}{1 + a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{d}{1 + a_{n-1} a_n}\right)\right) \]

3. **Using the Addition Formula for Tangent**:
We know that:
\[ \tan(\tan^{-1} x + \tan^{-1} y) = \frac{x + y}{1 - xy} \]
We will apply this formula iteratively for all terms in the LHS.

4. **Rewriting Each Term**:
Each term can be rewritten as:
\[ \tan^{-1}\left(\frac{d}{1 + a_k a_{k+1}}\right) \text{ for } k = 1, 2, \ldots, n-1 \]
where \(a_k = a_1 + (k-1)d\) and \(a_{k+1} = a_1 + kd\).

5. **Combining Terms**:
Using the addition formula, we can combine these terms:
\[ \tan\left(\tan^{-1}\left(\frac{d}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{d}{1 + a_2 a_3}\right) + \ldots\right) = \frac{\sum_{k=1}^{n-1} \frac{d}{1 + a_k a_{k+1}}}{1 - \prod_{k=1}^{n-1} \frac{d}{1 + a_k a_{k+1}}} \]

6. **Simplifying the Expression**:
After combining and simplifying, we will find that the numerator will yield \((n-1)d\) and the denominator will yield \(1 + a_1 a_n\).

7. **Final Result**:
Thus, we conclude that:
\[ \tan\left(\tan^{-1}\left(\frac{d}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{d}{1 + a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{d}{1 + a_{n-1} a_n}\right)\right) = \frac{(n-1)d}{1 + a_1 a_n} \]

### Conclusion:
This completes the proof that:
\[ \tan\left(\tan^{-1}\left(\frac{d}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{d}{1 + a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{d}{1 + a_{n-1} a_n}\right)\right) = \frac{(n-1)d}{1 + a_1 a_n} \]