The value of $\cot\left(\sum_{n=1}^{50}\tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$ is

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Inverse Trigonometric Functions

The value of

cot (\sum_{n = 1}^{50} tan^{- 1} (\frac{1}{1 + n + n ^{2}}))

Answer: (A)

Note that

\frac{1}{1 + n + n ^{2}} = \frac{( n + 1 ) - n}{1 + n ( n + 1 )}

, so:

tan^{- 1} (\frac{1}{1 + n + n ^{2}}) = tan^{- 1} (n + 1) - tan^{- 1} (n)

Therefore:

n = 1 \sum 50 tan^{- 1} (\frac{1}{1 + n + n ^{2}}) = tan^{- 1} (51) - tan^{- 1} (1)

Using the formula

tan (A - B) = \frac{t a n A - t a n B}{1 + t a n A t a n B}

cot (tan^{- 1} (51) - tan^{- 1} (1)) = \frac{1}{tan ( tan ^{- 1} ( 51 ) - tan ^{- 1} ( 1 ))} = \frac{1 + 51 \cdot 1}{51 - 1} = \frac{52}{50} = \frac{26}{25}

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