Class 12: Inverse Trigonometric Functions – Exercise 4.11 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: Prove the following results:}$

$\displaystyle \text{(i) } \tan^{-1} \frac{1}{7}+ \tan^{-1} \frac{1}{13} = \tan^{-1} \frac{2}{9}$

$\displaystyle \text{(ii) } \sin^{-1} \frac{12}{13}+ \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{63}{16} = \pi$

$\displaystyle \text{(iii) } \tan^{-1} \frac{1}{4}+ \tan^{-1} \frac{2}{9} = \sin^{-1} \frac{1}{\sqrt{5}}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$

$\displaystyle \text{LHS} = \tan^{-1} \frac{1}{7}+ \tan^{-1} \frac{1}{13} \hspace{2.0cm} \Big[ \because \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle = \tan^{-1} \Big[ \frac{\frac{1}{7} + \frac{1}{13}}{1 - \frac{1}{7} \cdot \frac{1}{13}} \Big]$

$\displaystyle = \tan^{-1} \Big[ \frac{\frac{20}{19}}{\frac{90}{91}} \Big]$

$\displaystyle = \tan^{-1}\frac{2}{9} = \text{RHS}$

$\displaystyle \text{(ii)}$

$\displaystyle \text{LHS} = \sin^{-1} \frac{12}{13}+ \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{63}{16}$

$\displaystyle = \tan^{-1} \frac{\frac{12}{13}}{\sqrt{1 - \frac{144}{169}}} + \tan^{-1}\frac{ \sqrt{1 - \frac{16}{25} }}{\frac{4}{5}} + \tan^{-1} \frac{63}{16} \\ \\ {\hspace{2.0cm} \Big[ \because \sin^{-1} x = \tan^{-1} \frac{x}{\sqrt{1-x^2}} \text{ and } \cos^{-1}x = \tan^{-1} \frac{\sqrt{1-x^2}}{x} \Big] }$

$\displaystyle = \tan^{-1} \frac{\frac{12}{13}}{\frac{5}{13}} + \tan^{-1} \frac{\frac{3}{5}}{\frac{4}{5}}+ \tan^{-1} \frac{63}{16}$

$\displaystyle = \tan^{-1} \frac{12}{5} + \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{63}{16} \\ \\ {\hspace{2.0cm} \Big[ \because \tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy} \Big] }$

$\displaystyle = \pi + \tan^{-1} \Big[ \frac{ \frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \cdot \frac{3}{4}} \Big] + \tan^{-1} \frac{63}{16}$

$\displaystyle = \pi + \tan^{-1} \Big[ \frac{\frac{63}{20}}{\frac{-16}{20}} \Big] + \tan^{-1} \frac{63}{16}$

$\displaystyle = \pi + \tan^{-1} \Big[ \frac{-63}{16} \Big] + \tan^{-1} \frac{63}{16}$

$\displaystyle = \pi - \tan^{-1} \Big[ \frac{63}{16} \Big] + \tan^{-1} \frac{63}{16}$

$\displaystyle = \pi = \text{RHS}$

$\displaystyle \text{(iii)}$

$\displaystyle \text{LHS} = \tan^{-1} \frac{1}{4}+ \tan^{-1} \frac{2}{9}$

$\displaystyle = \tan^{-1} \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \cdot \frac{2}{9} }$

$\displaystyle = \tan^{-1} \frac{\frac{17}{36}}{\frac{34}{36}}$

$\displaystyle = \tan^{-1} \frac{1}{2}$

$\displaystyle = \sin^{-1} \frac{\frac{1}{2}}{\sqrt{1+ (\frac{1}{2})^2} }$

$\displaystyle = \sin^{-1} \frac{1}{\sqrt{5}} = \text{RHS}$

$\$

$\displaystyle \text{Question 2: Find the value of:} \tan^{-1} \frac{x}{y} - \tan^{-1} \frac{x-y}{x+y} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2011] }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We know } \Big[ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) , xy> -1 \Big]$

$\displaystyle \text{Now, } \tan^{-1} \frac{x}{y} - \tan^{-1} \frac{x-y}{x+y} = \tan^{-1} \Bigg\{ \frac{\frac{x}{y} - \frac{x-y}{x+y} }{1 -\frac{x}{y} \cdot \frac{x-y}{x+y} } \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{ \frac{x^2 + xy - xy + y^2}{y(x+y)} }{\frac{x^2 + y^2 + xy - xy}{y(x+y}} \Bigg\}$

$\displaystyle = \tan^{-1} (1)$

$\displaystyle = \tan^{-1} \Big(\tan \frac{\pi}{4} \Big)$

$\displaystyle = \frac{\pi}{4}$

$\displaystyle \text{Hence, } \tan^{-1} \Big( \frac{x}{y} \Big) - \tan^{-1} \Big(\frac{x-y}{x+y} \Big) = \frac{\pi}{4}$

$\$

$\displaystyle \text{Question 3: Solve the following equation for } x:$

$\displaystyle \text{(i) } \tan^{-1} 2x + \tan^{-1} 3x = n \pi + \frac{3 \pi}{4}$

$\displaystyle \text{(ii) } \tan^{-1} (x+1) + \tan^{-1} (x-1) = \tan^{-1} \frac{8}{31}$

$\displaystyle \text{(iii) } \tan^{-1} (x-1) + \tan^{-1} x + \tan^{-1} (x+1) = \tan^{-1} 3x$

$\displaystyle \text{(iv) } \tan^{-1} \Big( \frac{1-x}{1+x} \Big) - \frac{1}{2} \tan^{-1} x = 0 , \ \ \ \ \text{ where } x > 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2008, 2010, 2011] }$

$\displaystyle \text{(v) } \cot^{-1} x - \cot^{-1} (x+2)= \frac{\pi}{12}, \text{ where } x > 0$

$\displaystyle \text{(vi) } \tan^{-1} (x+2) + \tan^{-1} (x-2) = \tan^{-1} \frac{8}{79}, \ \ \ \ x> 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2010] }$

$\displaystyle \text{(vii) } \tan^{-1} \frac{x}{2} + \tan^{-1} \frac{x}{3} = \frac{\pi}{4} , 0 <x < \sqrt{6} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2010] }$

$\displaystyle \text{(viii) } \tan^{-1} \Big( \frac{x-2}{x-4} \Big) + \tan^{-1} \Big( \frac{x+2}{x+4} \Big) = \frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2014] }$

$\displaystyle \text{(ix) } \tan^{-1} (2+x) + \tan^{-1}(2-x) = \tan^{-1} \frac{2}{3} , \text{ where } x < -\sqrt{3} \text{ or } , x > \sqrt{3}$

$\displaystyle \text{(x) } \tan^{-1} \Big( \frac{x-2}{x-1}\Big)+ \tan^{-1} \Big( \frac{x+2}{x+1} \Big) = \frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2016] }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$

$\displaystyle \text{We know, } \Big[ \because \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \text{Given } \tan^{-1} 2x + \tan^{-1} 3x = n \pi + \frac{3 \pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{2x+3x}{1 - 2x \times 3x} \Big) = n \pi + \frac{3 \pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{5x}{1 - 6x^2} \Big) = n \pi + \frac{3 \pi}{4}$

$\displaystyle \Rightarrow \Big( \frac{5x}{1 - 6x^2} \Big) = \tan \Big( n \pi + \frac{3 \pi}{4} \Big)$

$\displaystyle \Rightarrow \frac{5x}{1 - 6x^2} = -1$

$\displaystyle \Rightarrow 5x = -1 + 6x^2$

$\displaystyle \Rightarrow 6x^2 - 5x -1 = 0$

$\displaystyle \Rightarrow (6x+1 ) (x - 1) = 0$

$\displaystyle \Rightarrow x = - \frac{1}{6} \ \ \ \ \ \ \text{[As x = 1 is not satisfying the equation] }$

$\displaystyle \text{(ii)}$

$\displaystyle \text{Given, }\tan^{-1} (x+1) + \tan^{-1} (x-1) = \tan^{-1} \frac{8}{31}$

$\displaystyle \text{We know, } \Big[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \Rightarrow \tan^{-1} \frac{(x+1) + ( x- 1)}{1 - ( x+1)(x-1) } = \tan^{-1} \frac{8}{31}$

$\displaystyle \Rightarrow \tan^{-1} \frac{2x}{(1 - x^2 + 1) } = \tan^{-1} \frac{8}{31}$

$\displaystyle \Rightarrow \frac{2x}{(1 - x^2 + 1) } = \frac{8}{31}$

$\displaystyle \Rightarrow \frac{2x}{(2 - x^2 ) } = \frac{8}{31} \ \ \ \ \ \ [ \text{Note: } x^2< 2 \Rightarrow -\sqrt{2} < x < \sqrt{2} ]$

$\displaystyle \Rightarrow 62x= 8 - 8x^2 + 8$

$\displaystyle \Rightarrow 8x^2 + 62x - 16 = 0$

$\displaystyle \Rightarrow 4x^2 + 31x - 8 = 0$

$\displaystyle \Rightarrow 4x(x+8)- 1(x+8) = 0$

$\displaystyle \Rightarrow (4x-1)(x+8) = 0$

$\displaystyle \Rightarrow x = -\frac{1}{4} \text{ or } x = - 8$

$\displaystyle \Rightarrow \text{Since, } -\sqrt{2} < x < \sqrt{2}, \text{ the root of the given equation is } x = \frac{1}{4}$

$\displaystyle \text{(iii)}$

$\displaystyle \text{We know, } \Big[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \text{and }$

$\displaystyle \text{We know, } \Big[ \tan^{-1} x - \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1+xy} \Big) \Big]$

$\displaystyle \text{Given, }\tan^{-1} (x-1) + \tan^{-1} x + \tan^{-1} (x+1) = \tan^{-1} 3x$

$\displaystyle \Rightarrow \tan^{-1} (x-1) + \tan^{-1} (x+1) = \tan^{-1} 3x - \tan^{-1} x$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{x+1 + x - 1}{1 - (x+1)(x-1)} \Big) = \tan^{-1} \Big( \frac{3x-x}{1 + 3x \times x} \Big)$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{2x}{2 - x^2} \Big) = \tan^{-1} \Big( \frac{2x}{1 + 3x^2} \Big)$

$\displaystyle \Rightarrow \frac{2x}{2 - x^2} = \frac{2x}{1 + 3x^2}$

$\displaystyle \Rightarrow 2-x^2 = 1 + 3x^2$

$\displaystyle \Rightarrow 4x^2 - 1 = 0$

$\displaystyle \Rightarrow (2x-1)(2x+1) = 0$

$\displaystyle \Rightarrow x = \pm \frac{1}{2}$

$\displaystyle \text{(iv)} {\hspace{10.0cm} \text{[CBSE 2008, 2010, 2011] } }$

$\displaystyle \text{Given, } \tan^{-1} \Big( \frac{1-x}{1+x} \Big) - \frac{1}{2} \tan^{-1} x = 0 , \ \ \ \ \text{ where } x > 0$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{1-x}{1+x} \Big) = \frac{1}{2} \tan^{-1} x$

$\displaystyle \Rightarrow \tan^{-1} 1 - \tan^{-1} x = \frac{1}{2} \tan^{-1} x$

$\displaystyle \Rightarrow \tan^{-1} 1 = \frac{3}{2} \tan^{-1} x$

$\displaystyle \Rightarrow \frac{\pi}{4} = \frac{3}{2} \tan^{-1} x$

$\displaystyle \Rightarrow \frac{\pi}{6} = \tan^{-1} x$

$\displaystyle \Rightarrow x = \frac{1}{\sqrt{3}}$

$\displaystyle \text{(v)}$

$\displaystyle \text{Given, } \cot^{-1} x - \cot^{-1} (x+2)= \frac{\pi}{12}$

$\displaystyle \Rightarrow \tan^{-1} \frac{1}{x} - \tan^{-1} \frac{1}{(x+2)}= \frac{\pi}{12}$

$\displaystyle \Rightarrow \tan^{-1} \frac{\frac{1}{x} - \frac{1}{x+2}}{1+ \frac{1}{x(x+2)}}= \frac{\pi}{12}$

$\displaystyle \Rightarrow \tan^{-1}\Bigg( \frac{\frac{2}{x(x+2)}}{\frac{x^2+2x+1}{x(x+2)}} \Bigg)= \frac{\pi}{12}$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{2}{x^2+2x+1} \Big)= \frac{\pi}{12}$

$\displaystyle \Rightarrow \Big( \frac{2}{x^2+2x+1} \Big)= \tan \frac{\pi}{12}$

$\displaystyle \Rightarrow \Big( \frac{2}{x^2+2x+1} \Big)= \tan \Big( \frac{\pi}{3} - \frac{\pi}{4} \Big)$

$\displaystyle \Rightarrow \Big( \frac{2}{x^2+2x+1} \Big) = \frac{\tan \frac{\pi}{3} - \tan \frac{\pi}{4}}{1 + \tan \frac{\pi}{3} \cdot \tan \frac{\pi}{4}}$

$\displaystyle \Rightarrow \Big( \frac{2}{x^2+2x+1} \Big) = \frac{\sqrt{3}-1}{\sqrt{3}+1}$

$\displaystyle \Rightarrow \Big( \frac{2}{x^2+2x+1} \Big) = \frac{2}{(\sqrt{3}+1)^2}$

$\displaystyle \Rightarrow x+1 = \sqrt{3}+1$

$\displaystyle \Rightarrow x= \sqrt{3}$

$\displaystyle \text{(vi)} {\hspace{12.0cm} \text{[CBSE 2010] } }$

$\displaystyle \Big[ \text{We know } \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \text{Given, } \tan^{-1} (x+2) + \tan^{-1} (x-2) = \tan^{-1} \frac{8}{79}$

$\displaystyle \Rightarrow \tan^{-1} \frac{x+2 + x-2}{1 - (x+2)(x-2)} = \tan^{-1} \frac{8}{79}$

$\displaystyle \Rightarrow \tan^{-1} \frac{2x}{5-x^2} = \tan^{-1} \frac{8}{79}$

$\displaystyle \Rightarrow \frac{2x}{5-x^2} = \frac{8}{79}$

$\displaystyle \Rightarrow 79x = 20 - 4x^2$

$\displaystyle \Rightarrow 4x^2 + 79x - 20 = 0$

$\displaystyle \Rightarrow 4x^2 + 80x - x - 20 = 0$

$\displaystyle \Rightarrow 4x( x+ 20) - 1(x+20)=0$

$\displaystyle \Rightarrow (4x-1)(x+20) = 0$

$\displaystyle \Rightarrow x = \frac{1}{4} \text{ or } x = - 20$

$\displaystyle \Rightarrow x = \frac{1}{4} \ \ \ \ [\because x > 0]$

$\displaystyle \text{(vii)} {\hspace{12.0cm} \text{[CBSE 2010] } }$

$\displaystyle \Big[ \text{We know } \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \text{Given, } \tan^{-1} \frac{x}{2} + \tan^{-1} \frac{x}{3} = \frac{\pi}{4} , 0 <x < \sqrt{6}$

$\displaystyle \Rightarrow \tan^{-1} \frac{\frac{x}{2} + \frac{x}{3}}{1 - \frac{x}{2} \cdot \frac{x}{3}}= \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{\frac{5x}{6}}{\frac{6-x^2}{6}} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{5x}{6-x^2} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{5x}{6-x^2}= \tan \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{5x}{6-x^2}= 1$

$\displaystyle \Rightarrow 5x = 6 - x^2$

$\displaystyle \Rightarrow x^2 + 5x - 6 = 0$

$\displaystyle \Rightarrow (x-1)(x+6) = 0$

$\displaystyle \Rightarrow x = 1 [\because 0 <x < \sqrt{6} ]$

$\displaystyle \text{(viii)} {\hspace{12.0cm} \text{[CBSE 2014] } }$

$\displaystyle \Big[ \text{We know } \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \text{Given, }\tan^{-1} \Big( \frac{x-2}{x-4} \Big) + \tan^{-1} \Big( \frac{x+2}{x+4} \Big) = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Bigg[ \frac{\frac{x-2}{x-4} +\frac{x+2}{x+4} }{1 - \frac{x-2}{x-4} \cdot\frac{x+2}{x+4}} \Bigg]= \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Bigg[ \frac{ \frac{x^2 + 2x - 8 + x^2 - 2x - 8}{(x-4)(x+4)} }{ \frac{x^2 - 16 - x^2 +4}{(x-4)(x+4)} } \Bigg]= \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{2x^2 - 16}{-12} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{2x^2 - 16}{-12} = \tan \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{2x^2 - 16}{-12} = 1$

$\displaystyle \Rightarrow 2x^2 = 4$

$\displaystyle \Rightarrow x^2 = 2$

$\displaystyle \Rightarrow x = \pm \sqrt{2}$

$\displaystyle \text{(ix)}$

$\displaystyle \Big[ \text{We know } \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \text{Given, } \tan^{-1} (2+x) + \tan^{-1}(2-x) = \tan^{-1} \frac{2}{3} , \text{ where } x < -\sqrt{3} \text{ or } , x > \sqrt{3}$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{2+x+2-x}{1 - (2+x)(2-x)} \Big) = \tan^{-1} \frac{2}{3}$

$\displaystyle \Rightarrow \tan^{-1} \frac{4}{1 - 4 + x^2} = \tan^{-1} \frac{2}{3}$

$\displaystyle \Rightarrow \frac{4}{1 - 4 + x^2} = \frac{2}{3}$

$\displaystyle \Rightarrow -6 + 2x^2 = 12$

$\displaystyle \Rightarrow x^2 = 9$

$\displaystyle \Rightarrow x = \pm 3$

$\displaystyle \text{(x)} {\hspace{12.0cm} \text{[CBSE 2016] } }$

$\displaystyle \Big[ \text{We know } \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1-xy} \Big) \Big]$

$\displaystyle \text{Given, } \tan^{-1} \Big( \frac{x-2}{x-1}\Big)+ \tan^{-1} \Big( \frac{x+2}{x+1} \Big) = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{ \frac{x-2}{x-1}+ \frac{x+2}{x+1} }{1 - \frac{x-2}{x-1} \cdot \frac{x+2}{x+1} } \Bigg)= \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{(x-2)(x+1) + (x-1)(x+2)}{(x-1)(x+1)- (x-2)(x+2)} \Bigg)= \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{x^2 - x - 2 + x^2 + x - 2}{(x^2-1)-(x^2-4)} \Bigg)= \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \Bigg( \frac{2x^2 -4}{3} \Bigg)= \frac{\pi}{4}$

$\displaystyle \Rightarrow 2x^2 - 4 = 3$

$\displaystyle \Rightarrow 2x^2 = 7$

$\displaystyle \Rightarrow x^2 = \frac{7}{2}$

$\displaystyle \Rightarrow x = \pm \sqrt{\frac{7}{2}}$

$\$

$\displaystyle \text{Question 4: Sum of the following series:} \\ \\ \tan^{-1} \frac{1}{3}+ \tan^{-1} \frac{2}{9} + \tan^{-1} \frac{4}{33} + \ldots + \tan^{-1} \frac{2^{n-1}}{1+2^{2n-1}}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } \tan^{-1} \frac{1}{3}+ \tan^{-1} \frac{2}{9} + \tan^{-1} \frac{4}{33} + \ldots + \tan^{-1} \frac{2^{n-1}}{1+2^{2n-1}}$

$\displaystyle \Rightarrow \tan^{-1} \Big( \frac{2 - 1}{1 + 2 \times 1} \Big) + \tan^{-1} \Big( \frac{4 - 2}{1 + 4 \times 2} \Big) + \tan^{-1} \Big( \frac{8 - 4}{1 + 8 \times 4} \Big) + \ldots + \tan^{-1} \Big( \frac{2^n - 2^{n-1}}{1 + 2^n \times 2^{2n-1}} \Big)$

$\displaystyle \Rightarrow (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 4 - \tan^{-1} 2) + (\tan^{-1} 8 - \tan^{-1} 4) + \ldots + (\tan^{-1} 2^{n-1} - \tan^{-1} 2^{n-2}) + (\tan^{-1} 2^n - \tan^{-1} 2^{n-1})$

$\displaystyle \Rightarrow \tan^{-1} 2^n -\tan^{-1} 1$

$\displaystyle \Rightarrow \tan^{-1} 2^n - \frac{\pi}{4}$

$\$

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Class 12: Inverse Trigonometric Functions – Exercise 4.11

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