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)`

Answer

N/a

Explanation

We have, `a_(1)=a,a_(2)=a+d,a_(3)=a+2d`
and `d = a_(2) - a_(1) = a_(3) - a_(2) = a_(4) - a_(3) = "……"= a_(n) - a_(n-1)`
Given that, `tan[tan^(-1)(d/(1+a_(1)a_(2)))+ tan^(-1)(d/(1+a_(2)a_(3))) + tan^(-1)((d)/(1+a_(3)a_(4)))+"....."+tan^(-1)((d)/(1+a_(n+1).a_(n-1)))]`
`=tan[tan^(-1)'(a_(2)-a_(1))/(1+a_(2).a_(1))+tan^(-1)'(a_(3)-a_(2))/(1+a_(3).a_(2))+"....."+tan^(-1)'(a_(n)-a_(n-1))/(1+a_(n).a_(n-1))]`
`= tan[(tan^(-1)a_(2)-tan^(-1)a_(1))+(tan^(-1)a_(3)-tan^(-1)a_(2))+"...."+(tan^(-1)a_(n)-tan^(-1)a_(n-1))]`
`= tan[tan^(-1)a_(n)-tan^(-1)a_(1)]`
`= tan[tan^(-1)'(a_(n)-a_(1))/(1+a_(n).a_(1))] , [:' tan^(-1)x-tan^(-1)y=tan^(-1)((x-y)/(1+xy))]`
`= (a_(n)-a_(1))/(1+a_(n).a_(1)), [:' tan(tan^(-1)x)=x]`