Match the following.

\begin{aligned} & \text{ (i) } \sum_{i=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 i^2}\right)=t \Rightarrow \sum_{i=1}^{\infty} \tan ^{-1}\left(\frac{2}{4 i^2-1+1}\right) \\ & \Rightarrow \sum_{i=1}^{\infty} \tan ^{-1}\left\{\frac{(2 i+1)-(2 i-1)}{1+(2 i-1)(2 i+1)}\right\} \\ & \Rightarrow\left(\tan ^{-1} 3-\tan ^{-1} 1\right)+\left(\tan ^{-1} 5-\tan ^{-1} 3\right)+\ldots+\left\{\tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)\right\} \\ & \therefore \quad \quad t=\lim _{n \rightarrow \infty}\left(\tan ^{-1}(2 n+1)-\tan ^{-1} 1\right)=\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{2 n}{1+2 n+1}\right)=\frac{\pi}{4} \\ & \therefore \quad \tan t=1 . \end{aligned} (ii) We have, $\cos \theta_1=\frac{1-\tan ^2 \frac{\theta_1}{2}}{1+\tan ^2 \frac{\theta_2}{2}}=\frac{a}{b+c}$ \Rightarrow \quad $\tan ^2$$\frac{\theta}{2}=\frac{b+c-a}{b+c+a}$ Also, $\quad \cos \theta_3=\frac{1-\tan ^2 \frac{\theta_3}{2}}{1+\tan ^2 \frac{\theta_3}{2}}=\frac{c}{a+b}$ \Rightarrow \quad $\tan ^2$$\frac{\theta_3}{2}=\frac{a+b-c}{a+b+c} \Rightarrow \tan ^2 \frac{\theta_1}{2}+\tan ^2 \frac{\theta_3}{2}=\frac{2 b}{a+b+c}=\frac{2 b}{3 b}=\frac{2}{3}, \quad$ [as, $a, b, c$ are in AP $\Rightarrow 2 b=a+c$ ] (iii) Line through $(0,1,0)$ and perpendicular to plane $x+2 y+2 z=0$ is given by $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-0}{2}=r \therefore P(r, 2 r+1,2 r)$ be the foot of perpendicular on the straight line, then $\begin{array}{ll} & r \cdot 1+(2 r+1) \cdot 2+(2 r) \cdot 2=0 \\ \Rightarrow \quad r= & \quad r\left(-\frac{2}{9}, \frac{5}{9},-\frac{4}{9}\right) \\ \therefore & \quad P(r) \\ \therefore & \text{ Required perpendicular distance }=\sqrt{\frac{4+25+16}{81}}=\frac{\sqrt{5}}{3} \text{ unit. } \end{array}$