Solveeit Logo
Login

Question

Mathematics Question on Inverse Trigonometric Functions

∑r=1∞\displaystyle \sum_{r=1}^\infty tan−1(11+r+r2)=........... tan^{-1}\left(\frac{1}{1+r+r^{2}}\right)=...........

A

π2\frac{\pi}{2}

B

π4\frac{\pi}{4}

C

2π3\frac{2\pi}{3}

D

None of these

Answer

π4\frac{\pi}{4}

Explanation

Solution

11+r+r2=11+r(r+1)=1r(r+1)1+1r(r+1)=1r−1r+11+1r.(1r+1)\frac{1}{1+r+r^{2}}=\frac{1}{1+r\left(r+1\right)}=\frac{\frac{1}{r\left(r+1\right)}}{1+\frac{1}{r\left(r+1\right)}}=\frac{\frac{1}{r}-\frac{1}{r+1}}{1+\frac{1}{r}.\left(\frac{1}{r+1}\right)} ∴tan−1(11+r+r2)=tan−11r−tan−11r+1\therefore tan^{-1}\left(\frac{1}{1+r+r^{2}}\right)=tan^{-1} \frac{1}{r}-tan^{-1} \frac{1}{r+1} ∴\therefore ∑r=1∞\displaystyle \sum_{r=1}^\infty tan−1(11+r+r2)=tan−11=π4tan^{-1}\left(\frac{1}{1+r+r^{2}}\right)=tan^{-1} 1=\frac{\pi}{4}