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 start by expressing the terms in the summation and then applying properties of the inverse tangent function.

### Step-by-Step Solution:

1. **Understanding the Sequence**:
Let \( a_1, a_2, a_3, \ldots, a_n \) be an arithmetic progression (A.P.) with the first term \( a_1 \) and common difference \( d \). Therefore, we can express the terms as:
\[
a_k = a_1 + (k-1)d \quad \text{for } k = 1, 2, \ldots, n.
...