[" If "a_(1),a_(2),a_(3),..." 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)))+...+((d)/(1+a_(n-1)a_(n)))}" is equal to "]

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)))]=

If a_(1),a_(2),a_(3),….a_(n) is 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)))} is equal to

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_(n-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) 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) are in arithmetic progression and d is the common diference, then tan^(-1)((d)/(1+a_(1)a_(2)))+tan^(-1)((d)/(1+a_(2)a_(3)))=

If a_(1), a_(2), a_(3) are in arithmetic progression and d is the common diference, then tan^(-1)((d)/(1+a_(1)a_(2)))+tan^(-1)((d)/(1+a_(2)a_(3)))=