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

We have
`tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_(n -1) a_(n)))`
`= tan^(-1) ((a_(2) -a_(1))/(1 + a_(1) a_(2))) + tan^(-1) ((a_(3) -a_(2))/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((a_(n) - a_(n -1))/(1 + a_(n -1) a_(n)))`
`= (tan^(-1) a_(2) - tan^(-1) a_(1)) + (tan^(-1) a_(3) - tan^(-1) a_(2)) + ...+ (tan^(-1) a_(n) - tan^(-1) a_(n))`
`= tan^(-1) a_(n) - tan^(-1) a_(1) = tan^(-1) ((a_(n) -a_(1))/(1 + a_(n) a_(1))) = tan^(-1) (((n -1))/(1 + a_(1) a_(n)))`
`rArr tan[tan^(-1) ((d)/(1 + a_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ... + tan^(-1) ((d)/(1 + a_(n -1) a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))`