JEE Main 2021 — Mathematics Trigonometry

$$\sum _{r=1}^{50}{\mathrm{tan}}^{-1}(\frac{1}{2{r}^{2}})=\sum _{r=1}^{50}{\mathrm{tan}}^{-1}(\frac{2}{4{r}^{2}})$$r=1∑50​tan−1(2r21​)=r=1∑50​tan−1(4r22​)

$$=\sum _{r=1}^{50}{\mathrm{tan}}^{-1}(\frac{2}{1+4{r}^{2}-1})$$=r=1∑50​tan−1(1+4r2−12​)

$$=\sum _{r=1}^{50}{\mathrm{tan}}^{-1}(\frac{2}{1+(2r-1)(2r+1)})$$=r=1∑50​tan−1(1+(2r−1)(2r+1)2​)

$$=\sum _{r=1}^{50}{\mathrm{tan}}^{-1}(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)})$$=r=1∑50​tan−1(1+(2r+1)(2r−1)(2r+1)−(2r−1)​)

$$=\sum _{r=1}^{50}[{\mathrm{tan}}^{-1}(2r+1)-{\mathrm{tan}}^{-1}(2r-1)]$$=r=1∑50​[tan−1(2r+1)−tan−1(2r−1)]

So, $\displaystyle \sum _{r=1}^{50}{\mathrm{tan}}^{-1}(\frac{1}{2{r}^{2}})={\mathrm{tan}}^{-1}(3)-{\mathrm{tan}}^{-1}(1)+{\mathrm{tan}}^{-1}(5)-{\mathrm{tan}}^{-1}(3)+{\mathrm{tan}}^{-1}(7)-{\mathrm{tan}}^{-1}(5)...........{\mathrm{tan}}^{-1}(101)-{\mathrm{tan}}^{-1}(99)={\mathrm{tan}}^{-1}(101)-{\mathrm{tan}}^{-1}(1)$r=1∑50​tan−1(2r21​)=tan−1(3)−tan−1(1)+tan−1(5)−tan−1(3)+tan−1(7)−tan−1(5)...........tan−1(101)−tan−1(99)=tan−1(101)−tan−1(1)

$$={\mathrm{tan}}^{-1}(\frac{101-1}{1+101})$$=tan−1(1+101101−1​)

$$={\mathrm{tan}}^{-1}(\frac{100}{102})=p$$=tan−1(102100​)=p

So, $\mathrm{tan}p=\frac{100}{102}=\frac{50}{51}$tanp=102100​=5150​