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

As `a_1,a_2,a_3...a_n` are in A.P.,

So, the left hand side can be written as,

`L.H.S. = tan[tan^-1((a_2-a_1)/(1+a_1a_2))+tan^-1((a_3-a_2)/(1+a_2a_3))+...+tan^-1((a_n-a_(n-1))/(1+a_(n-1)a_n))]`

`=tan[tan^-1a_2-tan^-1a_1+tan^-1a_3-tan^-1a_2+tan^-1a_4-tan^-1a_3...+tan^-1a_n-tan^-1a_(n-1)]`

`=tan(tan^-1(a_n) - tan^-1(a_1))`

`=tan(tan^-1((a_n-a_1)/(1+a_1a_n)))`

As, `a_n - a_1 = (n-1)*d`

So, it becomes,

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