If `a_(1), a_(2), a_(3), …..a_(n)` is an arithmetic progression with common difference d. Prove that `tan[tan^(-1)((d)/(1+a_(1)a_(2)))+tan^(-1)((d)/(1+a_(2)a_(3)))+…+tan^(-1)((d)/(1+a_(n)a_(n-1)))]=(a_(n)-a_(1))/(1+a_(1)a_(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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