Mathematics - Properties of Triangles Question with Solution | TestHub

$$\begin{array}{rl}& \text{ (i) }\sum _{i=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2i^2}\right)=t\Rightarrow \sum _{i=1}^{\infty }{\tan }^{-1}\left(\frac{2}{4i^2-1+1}\right)\\ & \Rightarrow \sum _{i=1}^{\infty }{\tan }^{-1}\left\{\frac{(2i+1)-(2i-1)}{1+(2i-1)(2i+1)}\right\}\\ & \Rightarrow \left({\tan }^{-1}3-{\tan }^{-1}1\right)+\left({\tan }^{-1}5-{\tan }^{-1}3\right)+\dots +\left\{{\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)\right\}\\ & \therefore t=\underset{n\to \infty }{\lim }\left({\tan }^{-1}(2n+1)-{\tan }^{-1}1\right)=\underset{n\to \infty }{\lim }{\tan }^{-1}\left(\frac{2n}{1+2n+1}\right)=\frac{\pi }{4}\\ & \therefore \tan t=1.\end{array}$$ (ii) We have, $\cos {\theta }_1=\frac{1-{\tan }^2\frac{{\theta }_1}{2}}{1+{\tan }^2\frac{{\theta }_2}{2}}=\frac{a}{b+c}$ $$\Rightarrow {\tan }^2\frac{\theta }{2}=\frac{b+c-a}{b+c+a}$$ Also, $\cos {\theta }_3=\frac{1-{\tan }^2\frac{{\theta }_3}{2}}{1+{\tan }^2\frac{{\theta }_3}{2}}=\frac{c}{a+b}$ $$\Rightarrow {\tan }^2\frac{{\theta }_3}{2}=\frac{a+b-c}{a+b+c}$$ $\Rightarrow {\tan }^2\frac{{\theta }_1}{2}+{\tan }^2\frac{{\theta }_3}{2}=\frac{2b}{a+b+c}=\frac{2b}{3b}=\frac{2}{3},$ [as, $a,b,c$ are in AP $\Rightarrow 2b=a+c$ ] (iii) Line through $(0,1,0)$ and perpendicular to plane $x+2y+2z=0$ is given by $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-0}{2}=r$ $\therefore P(r,2r+1,2r)$ be the foot of perpendicular on the straight line, then $$\begin{array}{ll}& r\cdot 1+(2r+1)\cdot 2+(2r)\cdot 2=0\\ \Rightarrow r=& r\left(-\frac{2}{9},\frac{5}{9},-\frac{4}{9}\right)\\ \therefore & P(r)\\ \therefore & \text{ Required perpendicular distance }=\sqrt{\frac{4+25+16}{81}}=\frac{\sqrt{5}}{3}\text{ unit. }\end{array}$$