wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If 100n=1tan1(n2+n+1)=aπbtan1c, where a,b,cN and a and b are relatively prime, then the value of acb is

Open in App
Solution

tan1(n2+n+1)
=tan1(n(n+1)+1(n+1)n)
=π2tan1((n+1)n1+n(n+1))
=π2(tan1(n+1)tan1n)

100n=1tan1(n2+n+1)
=100(π2)(tan1101tan11)=50π+π4tan1101
=201π4tan1(101)acb=2011014=25


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon