JEE Maths›Inverse Trigonometric Functions›Visual Solution
Visual SolutionTricky
Question
The value of ∑r=110tan⁡−1(11+r+r2)\displaystyle\sum_{r=1}^{10} \tan^{-1}\left(\dfrac{1}{1+r+r^2}\right) is:
(A)tan⁡−111−π4\tan^{-1}11 - \dfrac{\pi}{4}
(B)tan⁡−110−π4\tan^{-1}10 - \dfrac{\pi}{4}
(C)tan⁡−111\tan^{-1}11
(D)π2−tan⁡−111\dfrac{\pi}{2} - \tan^{-1}11
Solution Path
Decompose 1/(1+r+r^2) as ((r+1)-r)/(1+r(r+1)). Apply tan inverse difference formula to get telescoping series. Sum = tan inverse(11) - pi/4.
01Question Setup
1/4
Find ∑r=110tan⁡−1 ⁣(11+r+r2)\displaystyle\sum_{r=1}^{10} \tan^{-1}\!\left(\frac{1}{1+r+r^2}\right). Four options given.
Sum=  ?\text{Sum} = \;?
02Decompose the General Term
2/4
Key algebraic trick: 11+r+r2=(r+1)−r1+r(r+1)\frac{1}{1+r+r^2} = \frac{(r+1)-r}{1+r(r+1)}. By the tan⁡−1\tan^{-1} difference formula, each term becomes tan⁡−1(r+1)−tan⁡−1(r)\tan^{-1}(r+1) - \tan^{-1}(r).
tan⁡−1 ⁣11+r+r2=tan⁡−1(r+1)−tan⁡−1(r)\tan^{-1}\!\frac{1}{1+r+r^2} = \tan^{-1}(r+1) - \tan^{-1}(r)
03Telescope the SumKEY INSIGHT
3/4
Writing out all terms, intermediate values cancel pairwise. Only tan⁡−1(11)\tan^{-1}(11) and −tan⁡−1(1)-\tan^{-1}(1) survive.
Sum=tan⁡−1(11)−tan⁡−1(1)\text{Sum} = \tan^{-1}(11) - \tan^{-1}(1)
04Final Answer
4/4
Since tan⁡−1(1)=π/4\tan^{-1}(1) = \pi/4, the sum equals tan⁡−1(11)−π/4\tan^{-1}(11) - \pi/4.
tan⁡−111−π4\boxed{\tan^{-1}11 - \dfrac{\pi}{4}}
Concepts from this question1 concepts unlocked
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Telescoping tan inverse Series

TRICKY

Many series involving tan inverse telescope when each term is written as a difference. The key decomposition is: tan inverse(1/(1+r+r^2)) = tan inverse(r+1) - tan inverse(r), because (r+1-r)/(1+r(r+1)) = 1/(1+r+r^2). This makes the sum collapse to tan inverse(n+1) - tan inverse(1). Similar decompositions work for 1/(2r^2) and other patterns.

∑r=1ntan⁡−111+r+r2=tan⁡−1(n+1)−π4\sum_{r=1}^{n} \tan^{-1}\frac{1}{1+r+r^2} = \tan^{-1}(n+1) - \frac{\pi}{4}

Telescoping tan inverse series appear in JEE Main regularly. The trick is always the same: express the general term as a difference of two tan inverse values. Once you spot this pattern, the problem becomes trivial.

Sum to n terms of tan inverse seriesSum to infinity of tan inverse seriesProve series identity
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