JEE Main 2022 — Mathematics Trigonometry

Let $\displaystyle S=\sum _{r=1}^{n}{\mathrm{tan}}^{-1}(\frac{1}{{r}^{2}+3r+3})$S=r=1∑n​tan−1(r2+3r+31​)

So, ${T}_{r}={\mathrm{tan}}^{-1}\frac{1}{{r}^{2}+3r+2+1}$Tr​=tan−1r2+3r+2+11​

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

$$={\mathrm{tan}}^{-1}(r+2)-{\mathrm{tan}}^{-1}(r+1)$$=tan−1(r+2)−tan−1(r+1)

$${T}_{1}={\mathrm{tan}}^{-1}3-{\mathrm{tan}}^{-1}2$$T1​=tan−13−tan−12

$${T}_{2}={\mathrm{tan}}^{-1}4-{\mathrm{tan}}^{-1}3 . .$$T2​=tan−14−tan−13..

$${T}_{n}={\mathrm{tan}}^{-1}(n+2)-{\mathrm{tan}}^{-1}(n+1)$$Tn​=tan−1(n+2)−tan−1(n+1)

$$\Rightarrow \sum _{r=1}^{n}{\mathrm{tan}}^{-1}\frac{1}{{r}^{2}+3r+3}={\mathrm{tan}}^{-1}(n+2)-{\mathrm{tan}}^{-1}2$$⇒r=1∑n​tan−1r2+3r+31​=tan−1(n+2)−tan−12

i.e. $\displaystyle 6\mathrm{tan}(\underset{n\rightarrow \infty }{\mathrm{lim}}\sum _{r=1}^{n}{\mathrm{tan}}^{-1}\frac{1}{{r}^{2}+3r+3})=6\mathrm{tan}(\underset{n\rightarrow \infty }{\mathrm{lim}}\sum _{r=1}^{n}[{\mathrm{tan}}^{-1}(n+2)-{\mathrm{tan}}^{-1}2])$6tan(n→∞lim​r=1∑n​tan−1r2+3r+31​)=6tan(n→∞lim​r=1∑n​[tan−1(n+2)−tan−12])

$$=6\mathrm{tan}(\frac{\pi }{2}-{\mathrm{tan}}^{-1}2)$$=6tan(2π​−tan−12)

$$=6\mathrm{tan}({\mathrm{tan}}^{-1}\frac{1}{2})=3$$=6tan(tan−121​)=3