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_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_( - 1)a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))`

Explanation

To prove the given statement, we will follow the steps outlined in the video transcript and provide a detailed solution.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given an arithmetic progression (A.P.) \( a_1, a_2, a_3, \ldots, a_n \) with a common difference \( d \). We need to prove 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}
...