Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The mth term is ${\tan }^{-1}(1/[1+\mathrm{m}(\mathrm{m}+1)])={\tan }^{-1}(\mathrm{m}+1)-{\tan }^{-1}\mathrm{m}$. Thus ${\mathrm{S}}_{\mathrm{n}}={\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}1={\tan }^{-1}[\mathrm{n}/(\mathrm{n}+2)]$. This gives ${\mathrm{S}}_{10}={\tan }^{-1}(5/6),{\mathrm{S}}_{\infty }=\pi /4,{\mathrm{S}}_6={\tan }^{-1}(3/4)={\sin }^{-1}(3/5)$, not ${\sin }^{-1}(4/5)$, and ${\mathrm{S}}_{20}={\tan }^{-1}(10/11)={\cot }^{-1}(11/10)$.