By considering $\tan \left( A-B \right)$ or otherwise, show that \({{\tan }^{-1}}\left( k+1 \right

Question

By considering $\tan \left( A-B \right)$ or otherwise, show that ${{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}\left( k \right)={{\cot }^{-1}}\left( {{k}^{2}}+k+1 \right)$ for $k>0$.[2] Hence find $\sum\limits_{k=1}^{n}{{{\cot }^{-1}}\left( {{k}^{2}}+k+1 \right)}$ in terms of $n$.[3] Deduce the exact value of $\sum\limits_{k=1}^{\infty }{{{\cot }^{-1}}\left( {{k}^{2}}+k+1 \right)}$.[2]

Accepted Answer

(ii) $\tan^{-1}(n+1) - \frac{\pi}{4}$ (iii) $\frac{\pi}{4}$

2017 NYJC Promo Q5

2017 NYJC Promo Q5

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