The sum of the infinite series ^ -1 ((7)/(4))+ ^ -1 ((19)/(4))+ ^ -1 ((39)/(4))+ ^ -1 ((67)/(4))+ . is :-

Correct answer: D. (π)/(2)- ^ -1 ((1)/(2)) aligned & T_ n = ^ -1 ( 4 4 n^2+3 ) & T_ n = ^ -1 ( (n+(1)/(2))-(n-(1)/(2)) 1+(n+(1)/(2))(n-(1)/(2)) ) & T_ n = ^ -1 (n+(1)/(2))- ^ -1 (n-(1)/(2)) & T_1+T_2+ +T_ n = ^ -1 (n+(1)/(2))- ^ -1 ((1)/(2)) & S_ ∞ =(π)/(2)- ^ -1 ((1)/(2)) aligned option (4)

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Inverse Trigonometric Functions

The sum of the infinite series ^ -1 ((7)/(4))+ ^ -1 ((19)/(4))+ ^...

The sum of the infinite series

\cot ^{-1}

\left(\frac{7}{4}\right)+\cot ^{-1}

\left(\frac{19}{4}\right)+

\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots .

is :-

Options

A $\frac{\pi}{2}+\tan ^{-1}\left(\frac{1}{2}\right)$
B $\frac{\pi}{2}-\cot ^{-1}\left(\frac{1}{2}\right)$
C $\frac{\pi}{2}+\cot ^{-1}\left(\frac{1}{2}\right)$
D $\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)$

Correct answer

D. $\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)$

Step-by-step solution

\begin{aligned}<br/>& \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{4}{4 \mathrm{n}^2+3}\right) \\ & \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\left(\mathrm{n}+\frac{1}{2}\right)-\left(\mathrm{n}-\frac{1}{2}\right)}{1+\left(\mathrm{n}+\frac{1}{2}\right)\left(\mathrm{n}-\frac{1}{2}\right)}\right) \\ & \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\mathrm{n}-\frac{1}{2}\right) \\ & \mathrm{T}_1+\mathrm{T}_2+\ldots+\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \mathrm{S}_{\infty}=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)<br/>\end{aligned}

option (4)

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