JEE Main 2025 — Inverse Trigonometric Functions Question with Solution
JEE Main 2025 (4 Apr Shift 2)
Question
The sum of the infinite series
is :-
is :-
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
$\begin{aligned}
& \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{4}{4 \mathrm{n}^2+3}\right) \\ & \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\left(\mathrm{n}+\frac{1}{2}\right)-\left(\mathrm{n}-\frac{1}{2}\right)}{1+\left(\mathrm{n}+\frac{1}{2}\right)\left(\mathrm{n}-\frac{1}{2}\right)}\right) \\ & \mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\mathrm{n}-\frac{1}{2}\right) \\ & \mathrm{T}_1+\mathrm{T}_2+\ldots+\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\mathrm{n}+\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \mathrm{S}_{\infty}=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}$
option (4)
& \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)
\end{aligned}$
option (4)
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This is a previous-year question from JEE Main 2025, covering the Inverse Trigonometric Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.