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)`

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/(11+a_(n-1)a_n))]=((n-1)d)/(1+a_1a_n)

If a_1, a_2,a_3, ,a_n is an A.P. with common difference d , then "tan"[tan^(-1)(d/(1+a_1a_2))+tan^(-1)(d/(1+a_2a_3))+...+tan^(-1)(d/(1+a_(n-1)a_n))]=

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_(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))

If a_1,a_2,a_3,…….a_n is an AP with common difference d then tan[tan^(-1)(d/(1+a_1a_2))+tan^(-1)((d)/(1+a_2a_3))+...+tan^(-1)((d)/(1+a_(n-1)a_n))] is equal to a) ((n-1)d)/(a_1+a_n) b) ((n-1)d)/(1+a_1a_n) c) (nd)/(1+a_1a_n) d) (a_n-a_1)/(a_n+a_1)

If a_(1),a_(2),a_(3),....,a_(n) is an A.P. with common difference d, then 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)))=

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_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_( - 1)a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))

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_(1) a_(2))) + tan^(-1) ((d)/(1 + a_(2) a_(3))) + ...+ tan^(-1) ((d)/(1 + a_( - 1)a_(n)))] = ((n -1)d)/(1 + a_(1) a_(n))

If a_(1),a_(2),a_(3),...a_(n) are in A.P. with common difference d, then 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)))]=