The sum of series `sec^(-1)sqrt(2)+sec^(-1)(sqrt(10))/3+sec^(-1)(sqrt(50))/7++sec^(-1)sqrt(((n^2+1)(n^2-2n+2))/((n^2-n+1)^2))` is `tan^(-1)1` (b) `n` `tan^(-1)(n+1)` (d) `tan^(-1)(n-1)`

Explanation

Let `S = sec^-1sqrt2+sec^-1(sqrt10/3)+sec^-1(sqrt50/7)+...+sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2)`

Here, `T_n = sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2)`

Let `sec^-1sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2) = theta`

`=>sec theta =sqrt(((n^2+1)(n^2-2n+2))/(n^2-n+1)^2`

`=>sec^2theta = ((n^2+1)(n^2-2n+2))/(n^2-n+1)^2`

`=>sec^2theta = ((n^2+1)(n^2+1-2n+1))/(n^2-n+1)^2`

`=>1+tan^2theta = ((n^2+1)(n^2+1)-2n(n^2+1)+n^2+1)/(n^2-n+1)^2`

`=>1+tan^2theta = ((n^2+1-n)^2+1)/(n^2-n+1)^2`

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