Given expression& =tan−12+tan−192+tan−1252+tan−1492+........
General term=(2n−1)22=4n2−4n+12=1+4n(n−1)2=1+2n(2n−2)2n−(2n−2)
Tn=tan−12n−tan−1(2n−2)
∴ & Sum of the series =tan−12−tan−10+tan−14−tan−12+tan−16−tan−14+…tan−12n−tan−1(2n−2)
=tan−12n−tan−10=tan−12n