Solution

Since a₁, a₂, a₃, … , aₙ are in A.P., we have:

\[ a_2 – a_1 = a_3 – a_2 = \cdots = a_n – a_{n-1} = d \]

The given expression becomes:

\[ \tan\Big[ \tan^{-1}\!\left(\frac{a_2-a_1}{1+a_1a_2}\right) + \tan^{-1}\!\left(\frac{a_3-a_2}{1+a_2a_3}\right) + \cdots + \tan^{-1}\!\left(\frac{a_n-a_{n-1}}{1+a_{n-1}a_n}\right) \Big] \]

Using the identity:

\[ \tan^{-1}\!\left(\frac{x-y}{1+xy}\right) = \tan^{-1}x – \tan^{-1}y \]

We get:

\[ = \tan\Big[ (\tan^{-1}a_2 – \tan^{-1}a_1) + (\tan^{-1}a_3 – \tan^{-1}a_2) + \cdots + (\tan^{-1}a_n – \tan^{-1}a_{n-1}) \Big] \]

All intermediate terms cancel out, giving:

\[ = \tan\big(\tan^{-1}a_n – \tan^{-1}a_1\big) \]

Again using the identity:

\[ \tan(\tan^{-1}x – \tan^{-1}y) = \frac{x-y}{1+xy} \]

\[ \boxed{ \frac{a_n – a_1}{1 + a_n a_1} } \]