Verified Solution • 2025–26 Session moderate QJEEM23Ma993c: JEE Main 2023 Mathematics - Class 12 JEEM23Ma993c Solution limt→0(11sin2t+21sin2t + ... + n1sin2t)sin2t\mathop {\lim }\limits_{t \to 0} {\left( {{1^{{1 \over {{{\sin }^2}t}}}} + {2^{{1 \over {{{\sin }^2}t}}}}\, + \,...\, + \,{n^{{1 \over {{{\sin }^2}t}}}}} \right)^{{{\sin }^2}t}}t→0lim(1sin2t1+2sin2t1+...+nsin2t1)sin2t is equal to Multiple Choice Options A n(n+1)2{{n(n + 1)} \over 2}2n(n+1) B n C n 2^22 + n D n 2^22 Step-by-Step Verified Solution amp;limt→0(1cosec2t+2cosec2t+……..+ncosec2t)sin2tamp;=limt→0n((1n)cosec2t+(2n)cosec2t+……..+1)sin2tamp;=n\begin{aligned}& \lim _{t \rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots \ldots . .+n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t} \\\\& =\lim _{t \rightarrow 0} n\left(\left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t}+\left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t}+\ldots \ldots . .+1\right)^{\sin ^2 t} \\\\& =\mathrm{n}\end{aligned}amp;t→0lim(1cosec2t+2cosec2t+……..+ncosec2t)sin2tamp;=t→0limn((n1)cosec2t+(n2)cosec2t+……..+1)sin2tamp;=n Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring