1.limt→0(1sin2t+2sin2t+⋯+nsin2t)sin2t\displaystyle \lim_{t\to 0} \left( \frac{1}{\sin^2 t} + \frac{2}{\sin^2 t} + \dots + \frac{n}{\sin^2 t} \right) \sin^2 tt→0lim(sin2t1+sin2t2+⋯+sin2tn)sin2t is equal toLogin to continueOnly logged in users canattempt or see the solution.
