Class 12: Inverse Trigonometric Functions

$\displaystyle \textbf{Question 1: }\text{Prove the following results:}$
$\displaystyle \text{(i) }\tan^{-1}\frac17+\tan^{-1}\frac1{13}=\tan^{-1}\frac29$
$\displaystyle \text{(ii) }\sin^{-1}\frac{12}{13}+\cos^{-1}\frac45+\tan^{-1}\frac{63}{16}=\pi$
$\displaystyle \text{(iii) }\tan^{-1}\frac14+\tan^{-1}\frac29=\sin^{-1}\frac1{\sqrt5}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \text{LHS}=\tan^{-1}\frac17+\tan^{-1}\frac1{13}$
$\displaystyle =\tan^{-1}\left[\frac{\frac17+\frac1{13}}{1-\frac17\cdot\frac1{13}}\right]\qquad\left[\because\ \tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right),\ xy<1\right]$
$\displaystyle =\tan^{-1}\left[\frac{\frac{20}{91}}{\frac{90}{91}}\right]$
$\displaystyle =\tan^{-1}\frac29=\text{RHS}$
$\displaystyle \\$

$\displaystyle \text{(ii)}$
$\displaystyle \text{LHS}=\sin^{-1}\frac{12}{13}+\cos^{-1}\frac45+\tan^{-1}\frac{63}{16}$
$\displaystyle =\tan^{-1}\left(\frac{\frac{12}{13}}{\sqrt{1-\frac{144}{169}}}\right)+\tan^{-1}\left(\frac{\sqrt{1-\frac{16}{25}}}{\frac45}\right)+\tan^{-1}\frac{63}{16}$
$\displaystyle \qquad\left[\because\ \sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\text{ and }\cos^{-1}x=\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right]$
$\displaystyle =\tan^{-1}\frac{12}{5}+\tan^{-1}\frac34+\tan^{-1}\frac{63}{16}$
$\displaystyle =\pi+\tan^{-1}\left[\frac{\frac{12}{5}+\frac34}{1-\frac{12}{5}\cdot\frac34}\right]+\tan^{-1}\frac{63}{16}$
$\displaystyle \qquad\left[\because\ \frac{12}{5}>0,\ \frac34>0\text{ and }\frac{12}{5}\cdot\frac34>1\right]$
$\displaystyle =\pi+\tan^{-1}\left[\frac{\frac{63}{20}}{-\frac{16}{20}}\right]+\tan^{-1}\frac{63}{16}$
$\displaystyle =\pi+\tan^{-1}\left(-\frac{63}{16}\right)+\tan^{-1}\frac{63}{16}$
$\displaystyle =\pi-\tan^{-1}\frac{63}{16}+\tan^{-1}\frac{63}{16}$
$\displaystyle =\pi=\text{RHS}$
$\displaystyle \\$

$\displaystyle \text{(iii)}$
$\displaystyle \text{LHS}=\tan^{-1}\frac14+\tan^{-1}\frac29$
$\displaystyle =\tan^{-1}\left[\frac{\frac14+\frac29}{1-\frac14\cdot\frac29}\right]\qquad\left[\because\ \frac14\cdot\frac29<1\right]$
$\displaystyle =\tan^{-1}\left[\frac{\frac{17}{36}}{\frac{34}{36}}\right]$
$\displaystyle =\tan^{-1}\frac12$
$\displaystyle =\sin^{-1}\left[\frac{\frac12}{\sqrt{1+\left(\frac12\right)^2}}\right]$
$\displaystyle =\sin^{-1}\frac1{\sqrt5}=\text{RHS}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the value of }\tan^{-1}\frac{x}{y}-\tan^{-1}\frac{x-y}{x+y}.\qquad\text{[CBSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }u=\frac{x}{y}\text{ and }v=\frac{x-y}{x+y},\text{ where }y\neq0\text{ and }x+y\neq0.$
$\displaystyle \frac{u-v}{1+uv}=\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y}\cdot\frac{x-y}{x+y}}$
$\displaystyle =\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}$
$\displaystyle =\frac{\frac{x^2+y^2}{y(x+y)}}{\frac{x^2+y^2}{y(x+y)}}=1$
$\displaystyle \text{Also, }1+uv=\frac{x^2+y^2}{y(x+y)}.$
$\displaystyle \text{Case I: When }y(x+y)>0,$
$\displaystyle uv>-1.$
$\displaystyle \therefore\tan^{-1}\frac{x}{y}-\tan^{-1}\frac{x-y}{x+y}=\tan^{-1}(1)=\frac{\pi}{4}$
$\displaystyle \text{Case II: When }y(x+y)<0,$
$\displaystyle uv<-1.$
$\displaystyle \therefore\tan^{-1}\frac{x}{y}-\tan^{-1}\frac{x-y}{x+y}=-\pi+\tan^{-1}(1)$
$\displaystyle =-\pi+\frac{\pi}{4}=-\frac{3\pi}{4}$
$\displaystyle \therefore\tan^{-1}\frac{x}{y}-\tan^{-1}\frac{x-y}{x+y}=\begin{cases}\dfrac{\pi}{4},&y(x+y)>0,\\[4pt]-\dfrac{3\pi}{4},&y(x+y)<0.\end{cases}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following equations for }x:$
$\displaystyle \text{(i) }\tan^{-1}2x+\tan^{-1}3x=n\pi+\frac{3\pi}{4}$
$\displaystyle \text{(ii) }\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\frac{8}{31}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$
$\displaystyle \tan^{-1}2x+\tan^{-1}3x=n\pi+\frac{3\pi}{4}$
$\displaystyle \text{Taking tangent on both sides,}$
$\displaystyle \frac{2x+3x}{1-(2x)(3x)}=\tan\left(n\pi+\frac{3\pi}{4}\right)$
$\displaystyle \Rightarrow\frac{5x}{1-6x^2}=-1$
$\displaystyle \Rightarrow5x=-1+6x^2$
$\displaystyle \Rightarrow6x^2-5x-1=0$
$\displaystyle \Rightarrow(6x+1)(x-1)=0$
$\displaystyle \Rightarrow x=-\frac16\text{ or }x=1$
$\displaystyle \text{When }x=1,$
$\displaystyle \tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\left(\frac{2+3}{1-6}\right)$
$\displaystyle =\pi+\tan^{-1}(-1)=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
$\displaystyle \therefore n=0.$
$\displaystyle \text{When }x=-\frac16,$
$\displaystyle \tan^{-1}\left(-\frac13\right)+\tan^{-1}\left(-\frac12\right)$
$\displaystyle =\tan^{-1}\left[\frac{-\frac13-\frac12}{1-\frac16}\right]$
$\displaystyle =\tan^{-1}(-1)=-\frac{\pi}{4}$
$\displaystyle =-\pi+\frac{3\pi}{4}$
$\displaystyle \therefore n=-1.$
$\displaystyle \therefore x=1\text{ when }n=0,\text{ and }x=-\frac16\text{ when }n=-1.$

$\displaystyle \text{(ii)}$
$\displaystyle \tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\frac{8}{31}$
$\displaystyle \text{Since the right-hand side lies in }\left(-\frac{\pi}{2},\frac{\pi}{2}\right),\text{ we require }(x+1)(x-1)<1.$
$\displaystyle \Rightarrow x^2-1<1$
$\displaystyle \Rightarrow x^2<2$
$\displaystyle \Rightarrow-\sqrt2<x<\sqrt2$
$\displaystyle \therefore\tan^{-1}\left[\frac{(x+1)+(x-1)}{1-(x+1)(x-1)}\right]=\tan^{-1}\frac{8}{31}$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{2x}{2-x^2}\right)=\tan^{-1}\frac{8}{31}$
$\displaystyle \Rightarrow\frac{2x}{2-x^2}=\frac{8}{31}$
$\displaystyle \Rightarrow62x=16-8x^2$
$\displaystyle \Rightarrow8x^2+62x-16=0$
$\displaystyle \Rightarrow4x^2+31x-8=0$
$\displaystyle \Rightarrow(4x-1)(x+8)=0$
$\displaystyle \Rightarrow x=\frac14\text{ or }x=-8$
$\displaystyle \text{Since }-\sqrt2<x<\sqrt2,\text{ we reject }x=-8.$
$\displaystyle \therefore x=\frac14$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following equations for }x:$
$\displaystyle \text{(iii) }\tan^{-1}(x-1)+\tan^{-1}x+\tan^{-1}(x+1)=\tan^{-1}3x$
$\displaystyle \text{(iv) }\tan^{-1}\left(\frac{1-x}{1+x}\right)-\frac12\tan^{-1}x=0,\text{ where }x>0\qquad\text{[CBSE 2008, 2010, 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii)}$
$\displaystyle \tan^{-1}(x-1)+\tan^{-1}x+\tan^{-1}(x+1)=\tan^{-1}3x$
$\displaystyle \Rightarrow\tan^{-1}(x-1)+\tan^{-1}(x+1)=\tan^{-1}3x-\tan^{-1}x$
$\displaystyle \text{Taking tangent on both sides,}$
$\displaystyle \frac{(x-1)+(x+1)}{1-(x-1)(x+1)}=\frac{3x-x}{1+3x^2}$
$\displaystyle \Rightarrow\frac{2x}{2-x^2}=\frac{2x}{1+3x^2}$
$\displaystyle \Rightarrow2x(1+3x^2)=2x(2-x^2)$
$\displaystyle \Rightarrow2x\left(1+3x^2-2+x^2\right)=0$
$\displaystyle \Rightarrow2x(4x^2-1)=0$
$\displaystyle \Rightarrow2x(2x-1)(2x+1)=0$
$\displaystyle \Rightarrow x=0,\ \frac12,\ -\frac12$
$\displaystyle \text{For }x=0,\quad\tan^{-1}(-1)+\tan^{-1}0+\tan^{-1}1=0=\tan^{-1}0.$
$\displaystyle \text{For }x=\frac12,\quad\tan^{-1}\left(-\frac12\right)+\tan^{-1}\frac12=0,$
$\displaystyle \therefore\tan^{-1}\left(-\frac12\right)+\tan^{-1}\frac12+\tan^{-1}\frac32=\tan^{-1}\frac32.$
$\displaystyle \text{For }x=-\frac12,\quad\tan^{-1}\left(-\frac12\right)+\tan^{-1}\frac12=0,$
$\displaystyle \therefore\tan^{-1}\left(-\frac32\right)+\tan^{-1}\left(-\frac12\right)+\tan^{-1}\frac12=\tan^{-1}\left(-\frac32\right).$
$\displaystyle \therefore x=0,\ \pm\frac12$

$\displaystyle \text{(iv)}$
$\displaystyle \tan^{-1}\left(\frac{1-x}{1+x}\right)-\frac12\tan^{-1}x=0,\qquad x>0$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{1-x}{1+x}\right)=\frac12\tan^{-1}x$
$\displaystyle \Rightarrow\tan^{-1}1-\tan^{-1}x=\frac12\tan^{-1}x$
$\displaystyle \qquad\left[\because\ \tan^{-1}1-\tan^{-1}x=\tan^{-1}\left(\frac{1-x}{1+x}\right),\ x>-1\right]$
$\displaystyle \Rightarrow\tan^{-1}1=\frac32\tan^{-1}x$
$\displaystyle \Rightarrow\frac{\pi}{4}=\frac32\tan^{-1}x$
$\displaystyle \Rightarrow\tan^{-1}x=\frac{\pi}{6}$
$\displaystyle \Rightarrow x=\tan\frac{\pi}{6}=\frac1{\sqrt3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following equations for }x:$
$\displaystyle \text{(v) }\cot^{-1}x-\cot^{-1}(x+2)=\frac{\pi}{12},\text{ where }x>0$
$\displaystyle \text{(vi) }\tan^{-1}(x+2)+\tan^{-1}(x-2)=\tan^{-1}\frac{8}{79},\ x>0\qquad\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(v)}$
$\displaystyle \cot^{-1}x-\cot^{-1}(x+2)=\frac{\pi}{12}$
$\displaystyle \Rightarrow\tan^{-1}\frac1x-\tan^{-1}\frac1{x+2}=\frac{\pi}{12}\qquad\left[\because x>0\right]$
$\displaystyle \Rightarrow\tan^{-1}\left[\frac{\frac1x-\frac1{x+2}}{1+\frac1{x(x+2)}}\right]=\frac{\pi}{12}$
$\displaystyle \Rightarrow\tan^{-1}\left[\frac{\frac{2}{x(x+2)}}{\frac{x^2+2x+1}{x(x+2)}}\right]=\frac{\pi}{12}$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{2}{(x+1)^2}\right)=\frac{\pi}{12}$
$\displaystyle \Rightarrow\frac{2}{(x+1)^2}=\tan\frac{\pi}{12}$
$\displaystyle \Rightarrow\frac{2}{(x+1)^2}=\tan\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$
$\displaystyle \Rightarrow\frac{2}{(x+1)^2}=\frac{\sqrt3-1}{\sqrt3+1}$
$\displaystyle \Rightarrow\frac{2}{(x+1)^2}=\frac{2}{(\sqrt3+1)^2}$
$\displaystyle \Rightarrow(x+1)^2=(\sqrt3+1)^2$
$\displaystyle \text{Since }x>0,\ x+1>0.$
$\displaystyle \therefore x+1=\sqrt3+1$
$\displaystyle \Rightarrow x=\sqrt3$

$\displaystyle \text{(vi)}$
$\displaystyle \tan^{-1}(x+2)+\tan^{-1}(x-2)=\tan^{-1}\frac{8}{79},\qquad x>0$
$\displaystyle \text{Since the right-hand side lies in }\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$
$\displaystyle (x+2)(x-2)<1$
$\displaystyle \Rightarrow x^2-4<1$
$\displaystyle \Rightarrow x^2<5$
$\displaystyle \therefore\tan^{-1}\left[\frac{(x+2)+(x-2)}{1-(x+2)(x-2)}\right]=\tan^{-1}\frac{8}{79}$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{2x}{5-x^2}\right)=\tan^{-1}\frac{8}{79}$
$\displaystyle \Rightarrow\frac{2x}{5-x^2}=\frac{8}{79}$
$\displaystyle \Rightarrow158x=40-8x^2$
$\displaystyle \Rightarrow4x^2+79x-20=0$
$\displaystyle \Rightarrow4x^2+80x-x-20=0$
$\displaystyle \Rightarrow4x(x+20)-(x+20)=0$
$\displaystyle \Rightarrow(4x-1)(x+20)=0$
$\displaystyle \Rightarrow x=\frac14\text{ or }x=-20$
$\displaystyle \text{Since }x>0,\text{ we reject }x=-20.$
$\displaystyle \therefore x=\frac14$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following equations for }x:$
$\displaystyle \text{(vii) }\tan^{-1}\frac{x}{2}+\tan^{-1}\frac{x}{3}=\frac{\pi}{4},\ 0<x<\sqrt6\qquad\text{[CBSE 2010]}$
$\displaystyle \text{(viii) }\tan^{-1}\left(\frac{x-2}{x-4}\right)+\tan^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi}{4}\qquad\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(vii)}$
$\displaystyle \tan^{-1}\frac{x}{2}+\tan^{-1}\frac{x}{3}=\frac{\pi}{4},\qquad 0<x<\sqrt6$
$\displaystyle \text{Since }\frac{x}{2}\cdot\frac{x}{3}=\frac{x^2}{6}<1,$
$\displaystyle \tan^{-1}\left[\frac{\frac{x}{2}+\frac{x}{3}}{1-\frac{x}{2}\cdot\frac{x}{3}}\right]=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{\frac{5x}{6}}{\frac{6-x^2}{6}}\right)=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{5x}{6-x^2}\right)=\frac{\pi}{4}$
$\displaystyle \Rightarrow\frac{5x}{6-x^2}=\tan\frac{\pi}{4}=1$
$\displaystyle \Rightarrow5x=6-x^2$
$\displaystyle \Rightarrow x^2+5x-6=0$
$\displaystyle \Rightarrow(x-1)(x+6)=0$
$\displaystyle \Rightarrow x=1\text{ or }x=-6$
$\displaystyle \text{Since }0<x<\sqrt6,\text{ we reject }x=-6.$
$\displaystyle \therefore x=1$

$\displaystyle \text{(viii)}$
$\displaystyle \tan^{-1}\left(\frac{x-2}{x-4}\right)+\tan^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi}{4},\qquad x\neq\pm4$
$\displaystyle \text{Using }\tan^{-1}u+\tan^{-1}v=\tan^{-1}\left(\frac{u+v}{1-uv}\right),$
$\displaystyle \tan^{-1}\left[\frac{\frac{x-2}{x-4}+\frac{x+2}{x+4}}{1-\frac{x-2}{x-4}\cdot\frac{x+2}{x+4}}\right]=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\left[\frac{\frac{2x^2-16}{x^2-16}}{\frac{-12}{x^2-16}}\right]=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{2x^2-16}{-12}\right)=\frac{\pi}{4}$
$\displaystyle \Rightarrow\frac{2x^2-16}{-12}=\tan\frac{\pi}{4}=1$
$\displaystyle \Rightarrow2x^2-16=-12$
$\displaystyle \Rightarrow2x^2=4$
$\displaystyle \Rightarrow x^2=2$
$\displaystyle \Rightarrow x=\pm\sqrt2$
$\displaystyle \text{For }x=\pm\sqrt2,\quad\frac{x-2}{x-4}\cdot\frac{x+2}{x+4}=\frac{x^2-4}{x^2-16}=\frac17<1.$
$\displaystyle \therefore x=\pm\sqrt2$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following equations for }x:$
$\displaystyle \text{(ix) }\tan^{-1}(2+x)+\tan^{-1}(2-x)=\tan^{-1}\frac23,\text{ where }x<-\sqrt3\text{ or }x>\sqrt3$
$\displaystyle \text{(x) }\tan^{-1}\left(\frac{x-2}{x-1}\right)+\tan^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}\qquad\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(ix)}$
$\displaystyle \tan^{-1}(2+x)+\tan^{-1}(2-x)=\tan^{-1}\frac23$
$\displaystyle \text{Since }x<-\sqrt3\text{ or }x>\sqrt3,\text{ we have }x^2>3.$
$\displaystyle \therefore(2+x)(2-x)=4-x^2<1$
$\displaystyle \therefore\tan^{-1}\left[\frac{(2+x)+(2-x)}{1-(2+x)(2-x)}\right]=\tan^{-1}\frac23$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{4}{1-(4-x^2)}\right)=\tan^{-1}\frac23$
$\displaystyle \Rightarrow\tan^{-1}\left(\frac{4}{x^2-3}\right)=\tan^{-1}\frac23$
$\displaystyle \Rightarrow\frac{4}{x^2-3}=\frac23$
$\displaystyle \Rightarrow12=2x^2-6$
$\displaystyle \Rightarrow2x^2=18$
$\displaystyle \Rightarrow x^2=9$
$\displaystyle \Rightarrow x=\pm3$
$\displaystyle \text{Both }x=-3\text{ and }x=3\text{ satisfy the given condition.}$
$\displaystyle \therefore x=\pm3$

$\displaystyle \text{(x)}$
$\displaystyle \tan^{-1}\left(\frac{x-2}{x-1}\right)+\tan^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4},\qquad x\neq\pm1$
$\displaystyle \text{Let }u=\frac{x-2}{x-1}\text{ and }v=\frac{x+2}{x+1}.$
$\displaystyle \tan^{-1}u+\tan^{-1}v=\frac{\pi}{4}$
$\displaystyle \Rightarrow\tan\left(\tan^{-1}u+\tan^{-1}v\right)=\tan\frac{\pi}{4}$
$\displaystyle \Rightarrow\frac{u+v}{1-uv}=1$
$\displaystyle \Rightarrow\frac{\frac{x-2}{x-1}+\frac{x+2}{x+1}}{1-\frac{x-2}{x-1}\cdot\frac{x+2}{x+1}}=1$
$\displaystyle \Rightarrow\frac{(x-2)(x+1)+(x+2)(x-1)}{(x-1)(x+1)-(x-2)(x+2)}=1$
$\displaystyle \Rightarrow\frac{x^2-x-2+x^2+x-2}{(x^2-1)-(x^2-4)}=1$
$\displaystyle \Rightarrow\frac{2x^2-4}{3}=1$
$\displaystyle \Rightarrow2x^2-4=3$
$\displaystyle \Rightarrow2x^2=7$
$\displaystyle \Rightarrow x^2=\frac72$
$\displaystyle \Rightarrow x=\pm\sqrt{\frac72}$
$\displaystyle \text{For }x=\pm\sqrt{\frac72},$
$\displaystyle uv=\frac{x^2-4}{x^2-1}=\frac{\frac72-4}{\frac72-1}=-\frac15<1.$
$\displaystyle \text{Hence, no }\pi\text{-adjustment is required, and both values satisfy the original equation.}$
$\displaystyle \therefore x=\pm\sqrt{\frac72}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the sum of the following series:}$
$\displaystyle \tan^{-1}\frac13+\tan^{-1}\frac29+\tan^{-1}\frac4{33}+\cdots+\tan^{-1}\frac{2^{n-1}}{1+2^{2n-1}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }S=\tan^{-1}\frac13+\tan^{-1}\frac29+\tan^{-1}\frac4{33}+\cdots+\tan^{-1}\frac{2^{n-1}}{1+2^{2n-1}}.$
$\displaystyle \text{The general term is}$
$\displaystyle \tan^{-1}\left(\frac{2^{r-1}}{1+2^{2r-1}}\right),\qquad r=1,2,\ldots,n.$
$\displaystyle \text{Now,}$
$\displaystyle \frac{2^{r-1}}{1+2^{2r-1}}=\frac{2^r-2^{r-1}}{1+2^r\cdot2^{r-1}}.$
$\displaystyle \therefore\tan^{-1}\left(\frac{2^{r-1}}{1+2^{2r-1}}\right)$
$\displaystyle =\tan^{-1}\left(\frac{2^r-2^{r-1}}{1+2^r\cdot2^{r-1}}\right)$
$\displaystyle =\tan^{-1}2^r-\tan^{-1}2^{r-1}.$
$\displaystyle \therefore S=(\tan^{-1}2-\tan^{-1}1)+(\tan^{-1}4-\tan^{-1}2)$
$\displaystyle \qquad+(\tan^{-1}8-\tan^{-1}4)+\cdots+(\tan^{-1}2^n-\tan^{-1}2^{n-1}).$
$\displaystyle \text{On cancelling the common terms,}$
$\displaystyle S=\tan^{-1}2^n-\tan^{-1}1$
$\displaystyle =\tan^{-1}2^n-\frac{\pi}{4}$
$\displaystyle \\$