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/(1+a_(n-1)a_n))]=((n-1)d)/(1+a_1a_n)`

Explanation

To prove the statement

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

we start with the given information that \(a_1, a_2, a_3, \ldots, a_n\) is an arithmetic progression (A.P.) with a common difference \(d\).

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