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Mathematics Question on Trigonometric Equations

If ∑k−1ntan⁡−1(1k2+k+1)=tan⁡−1(θ)\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) , then θ\theta =

A

nn+2\frac{n}{n + 2}

B

nn+1\frac{n}{n + 1}

C

11

D

nn−1\frac{n}{n -1 }

Answer

nn+2\frac{n}{n + 2}

Explanation

Solution

Given, ∑k=1ntan⁡−1(1k2+k+1)=tan⁡−1θ\displaystyle\sum_{k=1}^{n} \tan ^{-1}\left(\frac{1}{k^{2}+k+1}\right)=\tan ^{-1} \theta
Now, ∑k=1ntan⁡−1(1k2+k+1)=∑k=1ntan⁡−1[(k+1)−k]1+k(k+1)\displaystyle\sum_{k=1}^{n} \tan ^{-1}\left(\frac{1}{k^{2}+k+1}\right)=\displaystyle\sum_{k=1}^{n} \tan ^{-1} \frac{[(k+1)-k]}{1+k(k+1)}
=∑k=1n[tan⁡−1(k+1)−tan⁡−1k]=\displaystyle\sum_{k=1}^{n}\left[\tan ^{-1}(k+1)-\tan ^{-1} k\right]
=tan⁡−12−tan⁡−11+tan⁡−13−tan⁡−12+tan⁡−14=\tan ^{-1} 2-\tan ^{-1} 1+\tan ^{-1} 3-\tan ^{-1} 2+\tan ^{-1} 4
−tan⁡−13+…+tan⁡−1(n+1)−tan⁡−1n-\tan ^{-1} 3+\ldots+\tan ^{-1}(n+1)-\tan ^{-1} n
=tan⁡−1(n+1)−tan⁡−1l=tan⁡−1[n+1−11+n+1]=\tan ^{-1}(n+1)-\tan ^{-1} l =\tan ^{-1}\left[\frac{n+1-1}{1+n+1}\right]
=tan⁡−1[n2+n]=\tan ^{-1}\left[\frac{n}{2+n}\right]
∴tan⁡−1(n2+n)=tan⁡−1θ\therefore \tan ^{-1}\left(\frac{n}{2+n}\right)=\tan ^{-1} \theta
⇒θ=nn+2\Rightarrow \theta=\frac{n}{n+2}