If S= cot -1(n2+n+1)+ cot -1(n2+3 n+3)+ cot -1(n2+5 n+7)+ ldots ldots ldots+ cot -1(n2+39 n+381), then tan S is equal to

Question

If S= cot -1(n2+n+1)+ cot -1(n2+3 n+3)+ cot -1(n2+5 n+7)+ ldots ldots ldots+ cot -1(n2+39 n+381), then tan S is equal to (A) (20/n2+20 n+1) (B) (n/401+20 n) (C) (n/n2+20 n+1) (D) (20/401+20 n)

Tardigrade Admin · Accepted Answer

Correct answer is (a) (20/n2+20 n+1)

Q. If $S = cot^{- 1} (n^{2} + n + 1) + cot^{- 1} (n^{2} + 3 n + 3) + cot^{- 1} (n^{2} + 5 n + 7) + \dots\dots\dots + cot^{- 1} (n^{2} + 39 n + 381)$ , then $tan S$ is equal to

$\frac{20}{n ^{2} + 20 n + 1}$

$\frac{n}{401 + 20 n}$

$\frac{n}{n ^{2} + 20 n + 1}$

$\frac{20}{401 + 20 n}$

Correct answer is (a) $\frac{20}{n ^{2} + 20 n + 1}$

Q. If S=cot−1(n2+n+1)+cot−1(n2+3n+3)+cot−1(n2+5n+7)+………+cot−1(n2+39n+381), then tanS is equal to

n2+20n+120​

401+20nn​

n2+20n+1n​

401+20n20​