26. Inverse Circular Functions Solutions

  1. Let tan1(1)=θ\tan^{-1}(-1) = \theta then tanθ=1.\tan\theta = -1. Also, π2<θ<π2-\frac{\pi}{2}< \theta < \frac{\pi}{2}

    The only value in this range which satisfied this equation is π4.-\frac{\pi}{4}.

  2. Let cot1(1)=θ,\cot^{-1}(-1) = \theta, then cotθ=1\cot\theta = -1 and 0<θ<π0<\theta<\pi

    θ=3π4\Rightarrow \theta = \frac{3\pi}{4}

  3. Let sin1(32)=θ\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \theta then sinθ=32\sin\theta = -\frac{\sqrt{3}}{2}

    Also, π2θπ2θ=π3-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2} \Rightarrow \theta = -\frac{\pi}{3}

  4. sin[π3sin1(12)]\sin\left[\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2}\right)\right]

    =sin[π3{sin112}]=\sin\left[\frac{\pi}{3} -\{-\sin^{-1}\frac{1}{2}\}\right]

    =sin[π3+π6]=1= \sin\left[\frac{\pi}{3} + \frac{\pi}{6}\right] = 1

  5. sin[cos1(12)]=sin[πcos112]\sin\left[\cos^{-1}\left(-\frac{1}{2}\right)\right] = \sin\left[\pi - \cos^{-1}\frac{1}{2}\right]

    =sin2π3=32= \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}

  6. sin[tan1(3)+cos132]\sin\left[\tan^{-1}(-\sqrt{3}) + \cos^{-1}\frac{-\sqrt{3}}{2}\right]

    =sin[π3+ππ6]= \sin\left[-\frac{\pi}{3} + \pi- \frac{\pi}{6}\right]

    =sinπ2=1= \sin\frac{\pi}{2} = 1

  7. Given expression is tan[12cos153]\tan\left[\frac{1}{2}\cos^{-1}\frac{\sqrt{5}}{3}\right]

    Let cos153=2θ\cos^{-1}\frac{\sqrt{5}}{3} = 2\theta then cos2θ=53\cos2\theta = \frac{\sqrt{5}}{3}

    1tan2θ1+tan2θ=53\Rightarrow \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{\sqrt{5}}{3}

    By componendo and dividendo 2tan2θ2=353+5\frac{2\tan^2\theta}{2} = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}

    tanθ=±(352)\Rightarrow \tan\theta = \pm\left(\frac{3 - \sqrt{5}}{2}\right)

    Given 02θπ0θπ20 \leq 2\theta \leq \pi \Rightarrow 0\leq \theta \leq \frac{\pi}{2}

    i.e. θ\theta lies in first quadrant. tanθ=352\Rightarrow \tan\theta = \frac{3 - \sqrt{5}}{2}

  8. Given expression is sin1(sin2π3)\sin^{-1}\left(\sin\frac{2\pi}{3}\right)

    Let sin1(sin2π3)=θ\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \theta

    then sinθ=sin2π3\sin\theta = \sin\frac{2\pi}{3} and π2θπ2-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}

    =sin(ππ3)=sinπ3θ=π3=\sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{3}

    sin1(sin2π3)=π3\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}

  9. Let sin132=θsinθ=32\sin^{-1}\frac{\sqrt{3}}{2} = \theta \Rightarrow \sin\theta = \frac{\sqrt{3}}{2}

    θ=π3\Rightarrow \theta = \frac{\pi}{3} π2θπ2-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}

  10. Let tan113=θ\tan^{-1}\frac{-1}{\sqrt{3}} =\theta

    then tanθ=13=tan(π3)\tan\theta = \frac{-1}{\sqrt{3}} =\tan\left(\frac{-\pi}{3}\right)

    θ=π3\Rightarrow \theta = -\frac{\pi}{3}

  11. Let cot1(3)=θcotθ=3\cot^{-1}(-\sqrt{3}) = \theta \Rightarrow \cot\theta = -\sqrt{3}

    cotθ=cot(ππ6)\cot\theta = \cot\left(\pi - \frac{\pi}{6}\right)

    θ=5π6\theta = \frac{5\pi}{6}

  12. Let cot1cot5π4=θcotθ=cot(π+π4)\cot^{-1}\cot\frac{5\pi}{4} = \theta \Rightarrow \cot\theta = \cot\left(\pi + \frac{\pi}{4}\right)

    cotθ=cotπ4θ=π4\cot\theta = \cot\frac{\pi}{4} \Rightarrow \theta = \frac{\pi}{4}

  13. Let tan1(tan3π4)=θ\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \theta

    tanθ=tan3π4=tan(ππ4)\Rightarrow \tan\theta = \tan\frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right)

    θ=π4\Rightarrow \theta = -\frac{\pi}{4}

  14. sin112+cos112=π6+π3=π2\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}

  15. Let tan1=θtanθ=34cosθ=45\tan^{-1} = \theta \Rightarrow \tan\theta = \frac{3}{4} \Rightarrow \cos\theta = \frac{4}{5}

  16. cos[cos1(32)+π6]\cos\left[\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6}\right]

    =cos[π6+π6]=cosπ3=12=\cos\left[\frac{\pi}{6} + \frac{\pi}{6}\right] = \cos \frac{\pi}{3} = \frac{1}{2}

  17. L.H.S. =2tan113+tan117= 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}

    =tan1213119+tan117= \tan^{-1}\frac{2*\frac{1}{3}}{1 - \frac{1}{9}} + \tan^{-1}\frac{1}{7}

    =tan134+tan117= \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7}

    =tan134+171328=tan11=π4== \tan^{-1}\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{28}} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  18. L.H.S. =tan113+tan117+tan115+tan118= \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8}

    =tan113+171121+tan115+181140= \tan^{-1}\frac{\frac{1}{3} + \frac{1}{7}}{1 - \frac{1}{21}} + \tan^{-1}\frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{40}}

    =tan112+tan113=tan11=π4== \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  19. L.H.S. =sin145+sin1513+sin11665= \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65}

    =sin1(45125169+51311625)+sin11665= \sin^{-1}\left(\frac{4}{5}\sqrt{1 - \frac{25}{169}} + \frac{5}{13}\sqrt{1 - \frac{16}{25}}\right) + \sin^{-1}\frac{16}{65}

    =sin1(4865+1565)+sin11665=sin16465+sin11665= \sin^{-1}\left(\frac{48}{65} + \frac{15}{65}\right) + \sin^{-1}\frac{16}{65} = \sin^{-1}\frac{64}{65} + \sin^{-1}\frac{16}{65}

    =sin1(6365.1162652+16651632652)= \sin^{-1}\left(\frac{63}{65}.\sqrt{1 - \frac{16^2}{65^2}} + \frac{16}{65}\sqrt{1 - \frac{63^2}{65^2}}\right)

    sin1632+162652=sin1=π2=\sin^{-1}\frac{63^2 + 16^2}{65^2} = \sin^{-1} = \frac{\pi}{2} = R.H.S.

  20. We have to prove that 4tan115tan1170+tan1199=π44\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{70} + \tan^{-1}\frac{1}{99} = \frac{\pi}{4}

    4tan115=π4+tan1170tan1199\Rightarrow 4\tan^{-1}\frac{1}{5} = \frac{\pi}{4} + \tan^{-1}\frac{1}{70} - tan^{-1}\frac{1}{99}

    L.H.S. =4tan115=2tan12.151125=2tan1512= 4\tan^{-1}\frac{1}{5} = 2\tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{25}} = 2\tan^{-1}\frac{5}{12}

    =tan12.512125144=tan1120119= \tan^{-1}\frac{2.\frac{5}{12}}{1 - \frac{25}{144}} = \tan^{-1}\frac{120}{119}

    tan1170tan1199=tan11701991+170.199\tan^{-1}\frac{1}{70} - \tan^{-1}\frac{1}{99} = \tan^{-1}\frac{\frac{1}{70} - \frac{1}{99}}{1 + \frac{1}{70}.\frac{1}{99}}

    =tan11239= \tan^{-1}\frac{1}{239}

    R.H.S. =tan11+tan11239=tan1120119= \tan^{-1}1 + \tan^{-1}\frac{1}{239} = \tan^{-1}\frac{120}{119}

    Thus, L.H.S. = R.H.S.

  21. We have to prove that cot19+cosec1414=π4\cot^{-1}9 + \cosec^{-1}\frac{\sqrt{41}}{4} = \frac{\pi}{4}

    cot19=tan119\cot^{-1}9 = \tan^{-1}\frac{1}{9}

    Let cosec1414=θcosecθ=414\cosec^{-1}\frac{\sqrt{41}}{4} = \theta \Rightarrow \cosec\theta = \frac{\sqrt{41}}{4}

    Since we have to consider principal values only π2θπ2-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2} and θ0\theta \neq 0

    As cosecθ\cosec\theta is +ve here, θ\theta lies between 00 and π/2,\pi/2, hence tanθ\tan\theta must be positive.

    tanθ=45\Rightarrow \tan\theta = \frac{4}{5}

    tan119+tan145=tan119+4511945\tan^{-1}\frac{1}{9} + \tan^{-1}\frac{4}{5} = \tan^{-1}\frac{\frac{1}{9} + \frac{4}{5}}{1 - \frac{1}{9}\frac{4}{5}}

    =tan14145.4541=tan11=π4==\tan^{-1}\frac{41}{45}.\frac{45}{41} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  22. We have to prove that 4(cot13+cosec15)=π4(\cot^{-1}3 + \cosec^{-1}\sqrt{5}) = \pi

    cot13=tan113\cot^{-1}3 = \tan^{-1}\frac{1}{3}

    cosec15=tan112\cosec^{-1}\sqrt{5} = \tan^{-1}\frac{1}{2}

    L.H.S. =4(tan113+tan112)=4(tan112+13112.13)=4(\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{2}) = 4\left(\tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}.\frac{1}{3}}\right)

    =4tan11=π== 4\tan^{-1}1 = \pi = R.H.S.

  23. We have to prove that tan1x=2tan1[cosectan1xtancot1x]\tan^{-1}x = 2\tan^{-1}[\cosec\tan^{-1}x - \tan\cot^{-1}x]

    R.H.S. =2tan1[coseccosec11+x2xtantan11x]= 2\tan^{-1}[\cosec\cosec^{-1}\frac{\sqrt{1 + x^2}}{x} - \tan\tan^{-1}\frac{1}{x}]

    =2tan1(1+x21x)= 2\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)

    Let x=tanθ,x = \tan\theta, then R.H.S. =2tan1(secθ1tanθ)= 2\tan^{-1}\left(\frac{sec\theta - 1}{\tan\theta}\right)

    =2tan1(1cosθsinθ)=2tan1(2sin2θ22sinθ2cosθ2)=2\tan^{-1}\left(\frac{1 - \cos\theta}{\sin\theta}\right) = 2\tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)

    =2tan1.tanθ2=θ=tan1x== 2\tan^{-1}.\tan\frac{\theta}{2} = \theta = tan^{-1}x = L.H.S.

  24. 0<baaba+b\because 0 < b \leq a \therefore \sqrt{\frac{a - b}{a + b}} is real.

    L.H.S. =2tan1[aba+btanx2]= 2\tan^{-1}\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right]

    =cos1[1aba+btan2x21+aba+btan2x2]= \cos^{-1}\left[\frac{1 - \frac{a- b}{a + b}\tan^2\frac{x}{2}}{1 + \frac{a - b}{a + b}\tan^2\frac{x}{2}}\right]

    =cos1[a(1tan2x2)+b(1+tan2x2)a(1+tan2x2)+b(1+tan2x2)]= \cos^{-1}\left[\frac{a\left(1 - \tan^2\frac{x}{2}\right)+ b\left(1 + \tan^2\frac{x}{2}\right)}{a\left(1 + \tan^2\frac{x}{2}\right)+ b\left(1 + \tan^2\frac{x}{2}\right)}\right]

    =cos1[a(1tan2x21+tan2x2+b)a+b1tan2x21+tan2x2]= \cos^{-1}\left[\frac{a\left(\frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2} + b}\right)}{a + b\frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}}\right]

    =cos1[b+acosxa+bcosx]== \cos^{-1}\left[\frac{b + a\cos x}{a + b\cos x}\right] = R.H.S.

  25. L.H.S =tan1xy1+xy+tan1yz1+yz+tan1zx1+zx= \tan^{-1}\frac{x - y}{1 + xy} + \tan^{-1}\frac{y - z}{1 + yz} + \tan^{-1}\frac{z - x}{1 + zx}

    =tan1xtan1y+tan1ytan1z+tan1ztan1x=0= \tan^{-1}x - \tan^{-1}y + \tan^{-1}y - \tan^{-1}z + \tan^{-1}z - \tan^{-1}x = 0

    R.H.S. =tan1(x2y21+x2y2)+tan1(y2z21+y2z2)+tan1(z2x21+z2x2)= \tan^{-1}\left(\frac{x^2 - y^2}{1 + x^2y^2}\right) + \tan^{-1}\left(\frac{y^2 - z^2}{1 + y^2z^2}\right) + \tan^{-1}\left(\frac{z^2 - x^2}{1 + z^2x^2}\right)

    =tan1x2tan1y2+tan1y2tan1z2+tan1z2tan1x2=0= \tan^{-1}x^2 - \tan^{-1}y^2 + \tan^{-1}y^2 - \tan^{-1}z^2 + \tan^{-1}z^2 - \tan^{-1}x^2 = 0

    \therefore L.H.S. = R.H.S.

  26. We have to prove that sincot1tancos1x=x\sin\cot^{-1}\tan\cos^{-1}x = x

    L.H.S. =sincot1tantan11x2x= \sin\cot^{-1}\tan\tan^{-1}\frac{\sqrt{1 - x^2}}{x}

    =sincot11x2x= \sin\cot^{-1}\frac{\sqrt{1 - x^2}}{x}

    Let cot11x2x=θ\cot^{-1}\frac{1 - x^2}{x} = \theta then cotθ=1x2x\cot\theta = \frac{\sqrt{1 - x^2}}{x}

    sinθ=x\sin\theta = x

    Thus, L.H.S. = R.H.S.

  27. Case I: When π4<x<π2\frac{\pi}{4}<x<\frac{\pi}{2}

    0<cotx<10<\cot x< 1 and 0<cot3x<10<cotxcot3x<10<\cot^3x<1 \therefore 0<\cot x\cot^3x<1

    tan1cotx+tan1cot3x=tan1cotx+cot3x1cotxcot3x\tan^{-1}\cot x + \tan^{-1}\cot^3x =\tan^{-1}\frac{\cot x + \cot^3x}{1 - \cot x\cot^3x}

    =tan1cotx1cot2x=tan1tanxtan21=\tan^{-1}\frac{\cot x}{1 - \cot^2x} = \tan^{-1}\frac{\tan x}{\tan^2 - 1}

    =tan1(12tan2x)= -\tan^{-1}\left(\frac{1}{2}\tan 2x\right)

    tan1(12tan2x)+tan1cotx+tan1cot3x=0\Rightarrow \tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}\cot x + \tan^{-1}\cot^3x = 0

    Case II: When 0<x<π40<x<\frac{\pi}{4}

    cotx>1\cot x> 1 and cot3x>1\cot^3x > 1

    tan1cotx+tan1cot3x=πtan1(12tan2x)\Rightarrow \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi - \tan^{-1}\left(\frac{1}{2}\tan2x\right)

    tan1(12tan2x)+tan1cotx+tan1cot3x=π\Rightarrow \tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi

  28. tan112+tan113=tan112+13112.13\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}.\frac{1}{3}}

    =tan15/65/6=tan11=π4= \tan^{-1}\frac{5/6}{5/6} = \tan^{-1}1 = \frac{\pi}{4}

    tan135+tan114=tan135+14135.14\tan^{-1}\frac{3}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{\frac{3}{5} + \frac{1}{4}}{1 - \frac{3}{5}.\frac{1}{4}}

    =tan117/2017/20=tan11=π4=\tan^{-1}\frac{17/20}{17/20} = \tan^{-1}1 = \frac{\pi}{4}

  29. We have to prove that tan12ab3b+tan12ba3a=π3\tan^{-1}\frac{2a - b}{\sqrt{3}b} + \tan^{-1}\frac{2b - a}{\sqrt{3}a} = \frac{\pi}{3}

    L.H.S. =tan12ab3b+2ba3a1(2ab)(2ba)3ab= \tan^{-1}\frac{\frac{2a - b}{\sqrt{3}b} + \frac{2b - a}{\sqrt{3}a}}{1 - \frac{(2a - b)(2b - a)}{3ab}}

    =tan123a23ab+23b23ab3ab3ab4ab+2a2+2b2ab3ab= \tan^{-1}\frac{\frac{2\sqrt{3}a^2 - \sqrt{3}ab + 2\sqrt{3}b^2 -\sqrt{3}ab}{3ab}}{\frac{3ab - 4ab + 2a^2 + 2b^2 - ab}{3ab}}

    =tan12a2+23b223ab2a2+2b22ab=tan13=π3= \tan^{-1}\frac{2\sqrt{a^2} + 2\sqrt{3}b^2 - 2\sqrt{3}ab}{2a^2 + 2b^2 - 2ab} = \tan^{-1}\sqrt{3} = \frac{\pi}{3}

  30. We have to prove that tan125+tan113+tan1112=π4\tan^{-1}\frac{2}{5} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{12} = \frac{\pi}{4}

    L.H.S. =tan125+tan113+112113.112= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{\frac{1}{3} + \frac{1}{12}}{1 - \frac{1}{3}.\frac{1}{12}}

    =tan125+tan15/1235/36=tan125+tan137= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{5/12}{35/36} = \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{3}{7}

    =tan125+37125.37= \tan^{-1}\frac{\frac{2}{5} + \frac{3}{7}}{1 - \frac{2}{5}.\frac{3}{7}}

    =tan129/3529/35=tan11=π4= \tan^{-1}\frac{29/35}{29/35} = \tan^{-1}1 = \frac{\pi}{4}

  31. We have to prove that 2tan115+tan114=tan132432\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{32}{43}

    L.H.S. =tan12.151152+tan114= \tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{5^2}} + \tan^{-1}\frac{1}{4}

    =tan1512+tan114= \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{4}

    =tan1512+141512.14= \tan^{-1}\frac{\frac{5}{12} + \frac{1}{4}}{1 - \frac{5}{12}.\frac{1}{4}}

    =tan12/343/48=tan13243= \tan^{-1}\frac{2/3}{43/48} = \tan^{-1}\frac{32}{43}

  32. We have to prove that tan11+tan12+tan13=π=2(tan11+tan112+tan113)\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi = 2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)

    tan11+tan12+tan13=tan11+tan12+312.3=tan11+tan1(1)\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \tan^{-1}1 + \tan^{-1}\frac{2 + 3}{1 - 2.3} = \tan^{-1}1 + \tan^{-1}(-1)

    =tan1111+1=tan10=nπ= \tan^{-1}\frac{1 - 1}{1 + 1} =\tan^{-1}0 = n\pi

    2(tan11+tan112+tan113)2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)

    =2(tan11+tan112+12112.13)= 2\left(\tan^{-1}1 + \tan^{-1}\frac{\frac{1}{2} + \frac{1}{2}}{1 - \frac{1}{2}.\frac{1}{3}}\right)

    =2(π4+tan11)=2.π2=π= 2\left(\frac{\pi}{4} + \tan^{-1}1\right) = 2.\frac{\pi}{2} = \pi

    Thus, the above expression will have principal value as pi.pi.

  33. We have to prove that tan1x+cot1y=tan1xy+1yx\tan^{-1}x + \cot^{-1}y = \tan^{-1}\frac{xy + 1}{y - x}

    L.H.S. =tan1x+cot1y=tan1x+tan11y= \tan^{-1}x + \cot^{-1}y = \tan^{-1}x + \tan^{-1}\frac{1}{y}

    =tan1x+1y1x.1y=tan1xy+1yx= \tan^{-1}\frac{x + \frac{1}{y}}{1 - x.\frac{1}{y}} = \tan^{-1}\frac{xy + 1}{y - x}

  34. We have to prove that tan11x+y+tan1yx2+xy+1=cot1x\tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1} = \cot^{-1}x

    L.H.S. =tan11x+y+tan1yx2+xy+1= \tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1}

    =tan11x+y+yx2+xy+111x+y.yx2+xy+1= \tan^{-1}\frac{\frac{1}{x + y} + \frac{y}{x^2 + xy + 1}}{1 - \frac{1}{x + y}.\frac{y}{x^2 + xy + 1}}

    =tan1x2+2xy+y2+1x3+2x2y+xy2+x=tan11x=cot1x= \tan^{-1}\frac{x^2 + 2xy + y^2 + 1}{x^3 + 2x^2y + xy^2 + x} = \tan^{-1}\frac{1}{x} = \cot^{-1}x

  35. We have to prove that 2cot15+cot17+2cot18=π/42\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \pi/4

    We know that cot1x+cot1y=xy1x+y\cot^{-1}x + \cot^{-1}y = \frac{xy - 1}{x + y}

    2(cot15+cot18)=2cot13913=2cos13=cot143\therefore 2\left(\cot^{-1}5 + \cot^{-1}8\right) = 2\cot^{-1}\frac{39}{13} = 2\cos^{-1}3 = \cot^{-1}\frac{4}{3}

    2cot15+cot17+2cot18=cot143+cot17\therefore 2\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \cot^{-1}\frac{4}{3} + \cot^{-1}7

    =cot12831253=cot11=π/4=\cot^{-1}\frac{\frac{28}{3} - 1}{\frac{25}{3}} = \cot^{-1}1 = \pi/4

  36. We have to prove that tan1ab1+ab+tan1bc1+bc+tan1ca1+ca=0\tan^{-1}\frac{a - b}{1 + ab} + \tan^{-1}\frac{b - c}{1 + bc} + \tan^{-1}\frac{c - a}{1 + ca} = 0

    L.H.S. =tan1atan1b+tan1btan1c+tan1ctan1a=0= \tan^{-1}a - \tan^{-1}b + \tan^{-1}b - \tan^{-1}c + \tan^{-1}c - \tan^{-1}a = 0

  37. We have to prove that tan1a3b31+a3b3+tan1b3c31+b3c3+tan1c3a31+c3a3=0\tan^{-1}\frac{a^3 - b^3}{1 + a^3b^3} + \tan^{-1}\frac{b^3 - c^3}{1 + b^3c^3} + \tan^{-1}\frac{c^3 - a^3}{1 + c^3a^3} = 0

    L.H.S. =tan1a3tan1b3+tan1b3tan1c3+tan1c3tan1a3=0= \tan^{-1}a^3 - \tan^{-1}b^3 + \tan^{-1}b^3 - \tan^{-1}c^3 + \tan^{-1}c^3 - \tan^{-1}a^3 = 0

  38. We have to prove that cot1xy+1yx+cot1yz+1zy+cot1z=tan11x\cot^{-1}\frac{xy + 1}{y - x} + \cot^{-1}\frac{yz + 1}{z - y} + \cot^{-1}z = \tan^{-1}\frac{1}{x}

    L.H.S. =cot1xcot1y+cot1ycot1z+cot1z=cot1x=tan11x= \cot^{-1}x - \cot^{-1}y + \cot^{-1}y - \cot^{-1}z + \cot^{-1}z= \cot^{-1}x = \tan^{-1}\frac{1}{x}

  39. We have to prove that cos1(cosθ+cosϕ1+cosθcosϕ)=2tan1(tanθ2tanϕ2)\cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right) = 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)

    L.H.S. =cos1(cosθ+cosϕ1+cosθcosϕ)= \cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right)

    =tan11+cos2θcos2ϕ+2cosθcosϕcos2θcos2ϕ2cosθcosϕcosθ+cosϕ= \tan^{-1}\frac{\sqrt{1 + \cos^2\theta\cos^2\phi + 2\cos\theta\cos\phi - \cos^2\theta\cos^2\phi - 2\cos\theta\cos\phi}}{\cos\theta + \cos\phi}

    =tan1(1cos2θ)(1cos2ϕ)cosθ+cosϕ=tan1sinθsinϕcosθ+cosϕ= \tan^{-1}\frac{\sqrt{(1 - \cos^2\theta)(1 - \cos^2\phi)}}{\cos\theta + \cos\phi} = \tan^{-1}\frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}

    R.H.S. =2tan1(tanθ2tanϕ2)= 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)

    =tan12tanθ2tanϕ21tan2θ2tan2ϕ2= \tan^{-1}\frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}}{1 - \tan^2\frac{\theta}{2}\tan^2\frac{\phi}{2}}

    =tan12tanθ2tanϕ2.cos2θ2cos2ϕ2cos2θ2cos2ϕ2sin2θ2sin2ϕ2=\tan^{-1}\frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}.\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2}}{\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2} - \sin^2\frac{\theta}{2}\sin^2\frac{\phi}{2}}

    =tan112.sinθsinϕcos2θ2cos2ϕ2(1cos2θ2)(1cos2ϕ2)= \tan^{-1}\frac{1}{2}.\frac{\sin\theta\sin\phi}{\cos^2\frac{\theta}{2}\cos^2\frac{\phi}{2} - \left(1 - \cos^2\frac{\theta}{2}\right)\left(1-\cos^2\frac{\phi}{2}\right)}

    =tan1sinθsinϕcosθ+cosϕ= \tan^{-1}\frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}

  40. We have to prove that sin135+sin1817=sin17785\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}

    L.H.S. =sin135+sin1817= \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17}

    =sin1(35182172+81713252)=\sin^{-1}\left(\frac{3}{5}\sqrt{1 - \frac{8^2}{17^2}} + \frac{8}{17}\sqrt{1 - \frac{3^2}{5^2}}\right)

    =sin1(35.1517+817.45)= \sin^{-1}\left(\frac{3}{5}.\frac{15}{17} + \frac{8}{17}.\frac{4}{5}\right)

    sin1(45+3285)=sin17785=\sin^{-1}\left(\frac{45 + 32}{85}\right) = \sin^{-1}\frac{77}{85} = R.H.S.

  41. We have to prove that cos135+cos11213+cos16365=π2\cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65} = \frac{\pi}{2}

    We know that cos1x+cos1y=xy(1x2)(1y2)\cos^{-1}x + \cos^{-1}y = xy - \sqrt{(1 - x^2)(1 - y^2)}

    L.H.S. =cos135+cos11213+cos16365= \cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65}

    =cos1(35.1213(13252)(1122132))+cos16365= \cos^{-1}\left(\frac{3}{5}.\frac{12}{13} - \sqrt{\left(1 - \frac{3^2}{5^2}\right)\left(1 - \frac{12^2}{13^2}\right)}\right) + \cos^{-1}\frac{63}{65}

    =cos1(366545.513)+cos16365= \cos^{-1}\left(\frac{36}{65} - \frac{4}{5}.\frac{5}{13}\right) + \cos^{-1}\frac{63}{65}

    =cos1(36652065)+cos16365= \cos^{-1}\left(\frac{36}{65} - \frac{20}{65}\right) + \cos^{-1}\frac{63}{65}

    =cos11665+cos16365= \cos^{-1}\frac{16}{65} + \cos^{-1}\frac{63}{65}

    =cos1(1665.6364(1162652)(1632652))= \cos^{-1}\left(\frac{16}{65}.\frac{63}{64} - \sqrt{\left(1 - \frac{16^2}{65^2}\right)\left(1 - \frac{63^2}{65^2}\right)}\right)

    =cos10=π2== \cos^{-1}0 = \frac{\pi}{2} = R.H.S.

  42. We have to prove that sin1x+sin1y=cos1(1x21y2xy)\sin^{-1}x + \sin^{-1}y = \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right)

    L.H.S. =sin1x+sin1y=cos11x2+cos11y2= \sin^{-1}x + \sin^{-1}y = \cos^{-1}\sqrt{1 - x^2} + \cos^{-1}\sqrt{1 - y^2}

    =cos1(1x21y2[1(1x2)][1(1y2)])= \cos^{1-}(\sqrt{1 - x^2}\sqrt{1 - y^2} - \sqrt{[1 - (1 - x^2)][1 - (1 - y^2)]})

    =cos1(1x21y2xy)== \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right) = R.H.S.

  43. We have to prove that 4(sin1110+cos125)=π4\left(\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}}\right) =\pi

    or sin1110+cos125=π/4\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}} =\pi/4

    L.H.S. =sin1110+