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Question : 107 of 150
Marks:
+1,
-0
Solution:
We have,
tan−1() =[tan−1(m2+m+1) −tan−1(m2−m+1)] =(tan−13−tan−11)+(tan−17−tan−13) +(tan−1(13)−tan−17)+......+(tan−1(n2+n+1) −tan−1(n2−n+1)) =tan−1(n2+n+1)−tan−11 =tan−1() =tan−1()
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