26. Inverse Circular Functions Solutions

  1. Let tan⁡−1(−1)=θ\tan^{-1}(-1) = \theta then tan⁡θ=−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 cot⁡−1(−1)=θ,\cot^{-1}(-1) = \theta, then cot⁡θ=−1\cot\theta = -1 and 0<θ<π0<\theta<\pi

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

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

    Also, −π2≤θ≤π2⇒θ=−π3-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2} \Rightarrow \theta = -\frac{\pi}{3}

  4. sin⁡[π3−sin⁡−1(−12)]\sin\left[\frac{\pi}{3} -\sin^{-1}\left(-\frac{1}{2}\right)\right]

    =sin⁡[π3−{−sin⁡−112}]=\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⁡[cos⁡−1(−12)]=sin⁡[π−cos⁡−112]\sin\left[\cos^{-1}\left(-\frac{1}{2}\right)\right] = \sin\left[\pi - \cos^{-1}\frac{1}{2}\right]

    =sin⁡2π3=32= \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}

  6. sin⁡[tan⁡−1(−3)+cos⁡−1−32]\sin\left[\tan^{-1}(-\sqrt{3}) + \cos^{-1}\frac{-\sqrt{3}}{2}\right]

    =sin⁡[−π3+π−π6]= \sin\left[-\frac{\pi}{3} + \pi- \frac{\pi}{6}\right]

    =sin⁡π2=1= \sin\frac{\pi}{2} = 1

  7. Given expression is tan⁡[12cos⁡−153]\tan\left[\frac{1}{2}\cos^{-1}\frac{\sqrt{5}}{3}\right]

    Let cos⁡−153=2θ\cos^{-1}\frac{\sqrt{5}}{3} = 2\theta then cos⁡2θ=53\cos2\theta = \frac{\sqrt{5}}{3}

    ⇒1−tan⁡2θ1+tan⁡2θ=53\Rightarrow \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{\sqrt{5}}{3}

    By componendo and dividendo 2tan⁡2θ2=3−53+5\frac{2\tan^2\theta}{2} = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}

    ⇒tan⁡θ=±(3−52)\Rightarrow \tan\theta = \pm\left(\frac{3 - \sqrt{5}}{2}\right)

    Given 0≤2θ≤π⇒0≤θ≤π20 \leq 2\theta \leq \pi \Rightarrow 0\leq \theta \leq \frac{\pi}{2}

    i.e. θ\theta lies in first quadrant. ⇒tan⁡θ=3−52\Rightarrow \tan\theta = \frac{3 - \sqrt{5}}{2}

  8. Given expression is sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin\frac{2\pi}{3}\right)

    Let sin⁡−1(sin⁡2π3)=θ\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \theta

    then sin⁡θ=sin⁡2π3\sin\theta = \sin\frac{2\pi}{3} and −π2≤θ≤π2-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}

    =sin⁡(π−π3)=sin⁡π3⇒θ=π3=\sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{3}

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

  9. Let sin⁡−132=θ⇒sin⁡θ=32\sin^{-1}\frac{\sqrt{3}}{2} = \theta \Rightarrow \sin\theta = \frac{\sqrt{3}}{2}

    ⇒θ=π3\Rightarrow \theta = \frac{\pi}{3} −π2≤θ≤π2-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}

  10. Let tan⁡−1−13=θ\tan^{-1}\frac{-1}{\sqrt{3}} =\theta

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

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

  11. Let cot⁡−1(−3)=θ⇒cot⁡θ=−3\cot^{-1}(-\sqrt{3}) = \theta \Rightarrow \cot\theta = -\sqrt{3}

    cot⁡θ=cot⁡(π−π6)\cot\theta = \cot\left(\pi - \frac{\pi}{6}\right)

    θ=5π6\theta = \frac{5\pi}{6}

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

    cot⁡θ=cot⁡π4⇒θ=π4\cot\theta = \cot\frac{\pi}{4} \Rightarrow \theta = \frac{\pi}{4}

  13. Let tan⁡−1(tan⁡3π4)=θ\tan^{-1}\left(\tan\frac{3\pi}{4}\right) = \theta

    ⇒tan⁡θ=tan⁡3π4=tan⁡(π−π4)\Rightarrow \tan\theta = \tan\frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right)

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

  14. sin⁡−112+cos⁡−112=π6+π3=π2\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}

  15. Let tan⁡−1=θ⇒tan⁡θ=34⇒cos⁡θ=45\tan^{-1} = \theta \Rightarrow \tan\theta = \frac{3}{4} \Rightarrow \cos\theta = \frac{4}{5}

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

    =cos⁡[π6+π6]=cos⁡π3=12=\cos\left[\frac{\pi}{6} + \frac{\pi}{6}\right] = \cos \frac{\pi}{3} = \frac{1}{2}

  17. L.H.S. =2tan⁡−113+tan⁡−117= 2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}

    =tan⁡−12∗131−19+tan⁡−117= \tan^{-1}\frac{2*\frac{1}{3}}{1 - \frac{1}{9}} + \tan^{-1}\frac{1}{7}

    =tan⁡−134+tan⁡−117= \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7}

    =tan⁡−134+171−328=tan⁡−11=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. =tan⁡−113+tan⁡−117+tan⁡−115+tan⁡−118= \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8}

    =tan⁡−113+171−121+tan⁡−115+181−140= \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}}

    =tan⁡−112+tan⁡−113=tan⁡−11=π4== \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}1 = \frac{\pi}{4} = R.H.S.

  19. L.H.S. =sin⁡−145+sin⁡−1513+sin⁡−11665= \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65}

    =sin⁡−1(451−25169+5131−1625)+sin⁡−11665= \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}

    =sin⁡−1(4865+1565)+sin⁡−11665=sin⁡−16465+sin⁡−11665= \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}

    =sin⁡−1(6365.1−162652+16651−632652)= \sin^{-1}\left(\frac{63}{65}.\sqrt{1 - \frac{16^2}{65^2}} + \frac{16}{65}\sqrt{1 - \frac{63^2}{65^2}}\right)

    sin⁡−1632+162652=sin⁡−1=π2=\sin^{-1}\frac{63^2 + 16^2}{65^2} = \sin^{-1} = \frac{\pi}{2} = R.H.S.

  20. We have to prove that 4tan⁡−115−tan⁡−1170+tan⁡−1199=π44\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{70} + \tan^{-1}\frac{1}{99} = \frac{\pi}{4}

    ⇒4tan⁡−115=π4+tan⁡−1170−tan−1199\Rightarrow 4\tan^{-1}\frac{1}{5} = \frac{\pi}{4} + \tan^{-1}\frac{1}{70} - tan^{-1}\frac{1}{99}

    L.H.S. =4tan⁡−115=2tan⁡−12.151−125=2tan⁡−1512= 4\tan^{-1}\frac{1}{5} = 2\tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{25}} = 2\tan^{-1}\frac{5}{12}

    =tan⁡−12.5121−25144=tan⁡−1120119= \tan^{-1}\frac{2.\frac{5}{12}}{1 - \frac{25}{144}} = \tan^{-1}\frac{120}{119}

    tan⁡−1170−tan⁡−1199=tan⁡−1170−1991+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}}

    =tan⁡−11239= \tan^{-1}\frac{1}{239}

    R.H.S. =tan⁡−11+tan⁡−11239=tan⁡−1120119= \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 cot⁡−19+cosec⁡−1414=π4\cot^{-1}9 + \cosec^{-1}\frac{\sqrt{41}}{4} = \frac{\pi}{4}

    cot⁡−19=tan⁡−119\cot^{-1}9 = \tan^{-1}\frac{1}{9}

    Let cosec⁡−1414=θ⇒cosec⁡θ=414\cosec^{-1}\frac{\sqrt{41}}{4} = \theta \Rightarrow \cosec\theta = \frac{\sqrt{41}}{4}

    Since we have to consider principal values only −π2≤θ≤π2-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2} and θ≠0\theta \neq 0

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

    ⇒tan⁡θ=45\Rightarrow \tan\theta = \frac{4}{5}

    tan⁡−119+tan⁡−145=tan⁡−119+451−1945\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}}

    =tan⁡−14145.4541=tan⁡−11=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(cot⁡−13+cosec⁡−15)=π4(\cot^{-1}3 + \cosec^{-1}\sqrt{5}) = \pi

    cot⁡−13=tan⁡−113\cot^{-1}3 = \tan^{-1}\frac{1}{3}

    cosec⁡−15=tan⁡−112\cosec^{-1}\sqrt{5} = \tan^{-1}\frac{1}{2}

    L.H.S. =4(tan⁡−113+tan⁡−112)=4(tan⁡−112+131−12.13)=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)

    =4tan⁡−11=π== 4\tan^{-1}1 = \pi = R.H.S.

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

    R.H.S. =2tan⁡−1[cosec⁡cosec⁡−11+x2x−tan⁡tan⁡−11x]= 2\tan^{-1}[\cosec\cosec^{-1}\frac{\sqrt{1 + x^2}}{x} - \tan\tan^{-1}\frac{1}{x}]

    =2tan⁡−1(1+x2−1x)= 2\tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)

    Let x=tan⁡θ,x = \tan\theta, then R.H.S. =2tan⁡−1(secθ−1tan⁡θ)= 2\tan^{-1}\left(\frac{sec\theta - 1}{\tan\theta}\right)

    =2tan⁡−1(1−cos⁡θsin⁡θ)=2tan⁡−1(2sin⁡2θ22sin⁡θ2cos⁡θ2)=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)

    =2tan⁡−1.tan⁡θ2=θ=tan−1x== 2\tan^{-1}.\tan\frac{\theta}{2} = \theta = tan^{-1}x = L.H.S.

  24. ∵0<b≤a∴a−ba+b\because 0 < b \leq a \therefore \sqrt{\frac{a - b}{a + b}} is real.

    L.H.S. =2tan⁡−1[a−ba+btan⁡x2]= 2\tan^{-1}\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right]

    =cos⁡−1[1−a−ba+btan⁡2x21+a−ba+btan⁡2x2]= \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]

    =cos⁡−1[a(1−tan⁡2x2)+b(1+tan⁡2x2)a(1+tan⁡2x2)+b(1+tan⁡2x2)]= \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]

    =cos⁡−1[a(1−tan⁡2x21+tan⁡2x2+b)a+b1−tan⁡2x21+tan⁡2x2]= \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]

    =cos⁡−1[b+acos⁡xa+bcos⁡x]== \cos^{-1}\left[\frac{b + a\cos x}{a + b\cos x}\right] = R.H.S.

  25. L.H.S =tan⁡−1x−y1+xy+tan⁡−1y−z1+yz+tan⁡−1z−x1+zx= \tan^{-1}\frac{x - y}{1 + xy} + \tan^{-1}\frac{y - z}{1 + yz} + \tan^{-1}\frac{z - x}{1 + zx}

    =tan⁡−1x−tan⁡−1y+tan⁡−1y−tan⁡−1z+tan⁡−1z−tan⁡−1x=0= \tan^{-1}x - \tan^{-1}y + \tan^{-1}y - \tan^{-1}z + \tan^{-1}z - \tan^{-1}x = 0

    R.H.S. =tan⁡−1(x2−y21+x2y2)+tan⁡−1(y2−z21+y2z2)+tan⁡−1(z2−x21+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)

    =tan⁡−1x2−tan⁡−1y2+tan⁡−1y2−tan⁡−1z2+tan⁡−1z2−tan⁡−1x2=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 sin⁡cot⁡−1tan⁡cos⁡−1x=x\sin\cot^{-1}\tan\cos^{-1}x = x

    L.H.S. =sin⁡cot⁡−1tan⁡tan⁡−11−x2x= \sin\cot^{-1}\tan\tan^{-1}\frac{\sqrt{1 - x^2}}{x}

    =sin⁡cot⁡−11−x2x= \sin\cot^{-1}\frac{\sqrt{1 - x^2}}{x}

    Let cot⁡−11−x2x=θ\cot^{-1}\frac{1 - x^2}{x} = \theta then cot⁡θ=1−x2x\cot\theta = \frac{\sqrt{1 - x^2}}{x}

    sin⁡θ=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<cot⁡x<10<\cot x< 1 and 0<cot⁡3x<1∴0<cot⁡xcot⁡3x<10<\cot^3x<1 \therefore 0<\cot x\cot^3x<1

    tan⁡−1cot⁡x+tan⁡−1cot⁡3x=tan⁡−1cot⁡x+cot⁡3x1−cot⁡xcot⁡3x\tan^{-1}\cot x + \tan^{-1}\cot^3x =\tan^{-1}\frac{\cot x + \cot^3x}{1 - \cot x\cot^3x}

    =tan⁡−1cot⁡x1−cot⁡2x=tan⁡−1tan⁡xtan⁡2−1=\tan^{-1}\frac{\cot x}{1 - \cot^2x} = \tan^{-1}\frac{\tan x}{\tan^2 - 1}

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

    ⇒tan⁡−1(12tan⁡2x)+tan⁡−1cot⁡x+tan⁡−1cot⁡3x=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}

    cot⁡x>1\cot x> 1 and cot⁡3x>1\cot^3x > 1

    ⇒tan⁡−1cot⁡x+tan⁡−1cot⁡3x=π−tan⁡−1(12tan⁡2x)\Rightarrow \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi - \tan^{-1}\left(\frac{1}{2}\tan2x\right)

    ⇒tan⁡−1(12tan⁡2x)+tan⁡−1cot⁡x+tan⁡−1cot⁡3x=π\Rightarrow \tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}\cot x + \tan^{-1}\cot^3x = \pi

  28. tan⁡−112+tan⁡−113=tan⁡−112+131−12.13\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}}

    =tan⁡−15/65/6=tan⁡−11=π4= \tan^{-1}\frac{5/6}{5/6} = \tan^{-1}1 = \frac{\pi}{4}

    tan⁡−135+tan⁡−114=tan⁡−135+141−35.14\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}}

    =tan⁡−117/2017/20=tan⁡−11=π4=\tan^{-1}\frac{17/20}{17/20} = \tan^{-1}1 = \frac{\pi}{4}

  29. We have to prove that tan⁡−12a−b3b+tan⁡−12b−a3a=π3\tan^{-1}\frac{2a - b}{\sqrt{3}b} + \tan^{-1}\frac{2b - a}{\sqrt{3}a} = \frac{\pi}{3}

    L.H.S. =tan⁡−12a−b3b+2b−a3a1−(2a−b)(2b−a)3ab= \tan^{-1}\frac{\frac{2a - b}{\sqrt{3}b} + \frac{2b - a}{\sqrt{3}a}}{1 - \frac{(2a - b)(2b - a)}{3ab}}

    =tan⁡−123a2−3ab+23b2−3ab3ab3ab−4ab+2a2+2b2−ab3ab= \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}}

    =tan⁡−12a2+23b2−23ab2a2+2b2−2ab=tan⁡−13=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 tan⁡−125+tan⁡−113+tan⁡−1112=π4\tan^{-1}\frac{2}{5} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{12} = \frac{\pi}{4}

    L.H.S. =tan⁡−125+tan⁡−113+1121−13.112= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{\frac{1}{3} + \frac{1}{12}}{1 - \frac{1}{3}.\frac{1}{12}}

    =tan⁡−125+tan⁡−15/1235/36=tan⁡−125+tan⁡−137= \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{5/12}{35/36} = \tan^{-1}\frac{2}{5} + \tan^{-1}\frac{3}{7}

    =tan⁡−125+371−25.37= \tan^{-1}\frac{\frac{2}{5} + \frac{3}{7}}{1 - \frac{2}{5}.\frac{3}{7}}

    =tan⁡−129/3529/35=tan⁡−11=π4= \tan^{-1}\frac{29/35}{29/35} = \tan^{-1}1 = \frac{\pi}{4}

  31. We have to prove that 2tan⁡−115+tan⁡−114=tan⁡−132432\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{32}{43}

    L.H.S. =tan⁡−12.151−152+tan⁡−114= \tan^{-1}\frac{2.\frac{1}{5}}{1 - \frac{1}{5^2}} + \tan^{-1}\frac{1}{4}

    =tan⁡−1512+tan⁡−114= \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{4}

    =tan⁡−1512+141−512.14= \tan^{-1}\frac{\frac{5}{12} + \frac{1}{4}}{1 - \frac{5}{12}.\frac{1}{4}}

    =tan⁡−12/343/48=tan⁡−13243= \tan^{-1}\frac{2/3}{43/48} = \tan^{-1}\frac{32}{43}

  32. We have to prove that tan⁡−11+tan⁡−12+tan⁡−13=π=2(tan⁡−11+tan⁡−112+tan⁡−113)\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)

    tan⁡−11+tan⁡−12+tan⁡−13=tan⁡−11+tan⁡−12+31−2.3=tan⁡−11+tan⁡−1(−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)

    =tan⁡−11−11+1=tan⁡−10=nπ= \tan^{-1}\frac{1 - 1}{1 + 1} =\tan^{-1}0 = n\pi

    2(tan⁡−11+tan⁡−112+tan⁡−113)2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)

    =2(tan⁡−11+tan⁡−112+121−12.13)= 2\left(\tan^{-1}1 + \tan^{-1}\frac{\frac{1}{2} + \frac{1}{2}}{1 - \frac{1}{2}.\frac{1}{3}}\right)

    =2(π4+tan⁡−11)=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 tan⁡−1x+cot⁡−1y=tan⁡−1xy+1y−x\tan^{-1}x + \cot^{-1}y = \tan^{-1}\frac{xy + 1}{y - x}

    L.H.S. =tan⁡−1x+cot⁡−1y=tan⁡−1x+tan⁡−11y= \tan^{-1}x + \cot^{-1}y = \tan^{-1}x + \tan^{-1}\frac{1}{y}

    =tan⁡−1x+1y1−x.1y=tan⁡−1xy+1y−x= \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 tan⁡−11x+y+tan⁡−1yx2+xy+1=cot⁡−1x\tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1} = \cot^{-1}x

    L.H.S. =tan⁡−11x+y+tan⁡−1yx2+xy+1= \tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1}

    =tan⁡−11x+y+yx2+xy+11−1x+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}}

    =tan⁡−1x2+2xy+y2+1x3+2x2y+xy2+x=tan⁡−11x=cot⁡−1x= \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 2cot⁡−15+cot⁡−17+2cot⁡−18=π/42\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \pi/4

    We know that cot⁡−1x+cot⁡−1y=xy−1x+y\cot^{-1}x + \cot^{-1}y = \frac{xy - 1}{x + y}

    ∴2(cot⁡−15+cot⁡−18)=2cot⁡−13913=2cos⁡−13=cot⁡−143\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}

    ∴2cot⁡−15+cot⁡−17+2cot⁡−18=cot⁡−143+cot⁡−17\therefore 2\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \cot^{-1}\frac{4}{3} + \cot^{-1}7

    =cot⁡−1283−1253=cot⁡−11=π/4=\cot^{-1}\frac{\frac{28}{3} - 1}{\frac{25}{3}} = \cot^{-1}1 = \pi/4

  36. We have to prove that tan⁡−1a−b1+ab+tan⁡−1b−c1+bc+tan⁡−1c−a1+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. =tan⁡−1a−tan⁡−1b+tan⁡−1b−tan⁡−1c+tan⁡−1c−tan⁡−1a=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 tan⁡−1a3−b31+a3b3+tan⁡−1b3−c31+b3c3+tan⁡−1c3−a31+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. =tan⁡−1a3−tan⁡−1b3+tan⁡−1b3−tan⁡−1c3+tan⁡−1c3−tan⁡−1a3=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 cot⁡−1xy+1y−x+cot⁡−1yz+1z−y+cot⁡−1z=tan⁡−11x\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. =cot⁡−1x−cot⁡−1y+cot⁡−1y−cot⁡−1z+cot⁡−1z=cot⁡−1x=tan⁡−11x= \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 cos⁡−1(cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ)=2tan⁡−1(tan⁡θ2tan⁡ϕ2)\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. =cos⁡−1(cos⁡θ+cos⁡ϕ1+cos⁡θcos⁡ϕ)= \cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right)

    =tan⁡−11+cos⁡2θcos⁡2ϕ+2cos⁡θcos⁡ϕ−cos⁡2θcos⁡2ϕ−2cos⁡θcos⁡ϕcos⁡θ+cos⁡ϕ= \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}

    =tan⁡−1(1−cos⁡2θ)(1−cos⁡2ϕ)cos⁡θ+cos⁡ϕ=tan⁡−1sin⁡θsin⁡ϕcos⁡θ+cos⁡ϕ= \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. =2tan⁡−1(tan⁡θ2tan⁡ϕ2)= 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)

    =tan⁡−12tan⁡θ2tan⁡ϕ21−tan⁡2θ2tan⁡2ϕ2= \tan^{-1}\frac{2\tan\frac{\theta}{2}\tan\frac{\phi}{2}}{1 - \tan^2\frac{\theta}{2}\tan^2\frac{\phi}{2}}

    =tan⁡−12tan⁡θ2tan⁡ϕ2.cos⁡2θ2cos⁡2ϕ2cos⁡2θ2cos⁡2ϕ2−sin⁡2θ2sin⁡2ϕ2=\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}}

    =tan⁡−112.sin⁡θsin⁡ϕcos⁡2θ2cos⁡2ϕ2−(1−cos⁡2θ2)(1−cos⁡2ϕ2)= \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)}

    =tan⁡−1sin⁡θsin⁡ϕcos⁡θ+cos⁡ϕ= \tan^{-1}\frac{\sin\theta\sin\phi}{\cos\theta + \cos\phi}

  40. We have to prove that sin⁡−135+sin⁡−1817=sin⁡−17785\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}

    L.H.S. =sin⁡−135+sin⁡−1817= \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17}

    =sin⁡−1(351−82172+8171−3252)=\sin^{-1}\left(\frac{3}{5}\sqrt{1 - \frac{8^2}{17^2}} + \frac{8}{17}\sqrt{1 - \frac{3^2}{5^2}}\right)

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

    sin⁡−1(45+3285)=sin⁡−17785=\sin^{-1}\left(\frac{45 + 32}{85}\right) = \sin^{-1}\frac{77}{85} = R.H.S.

  41. We have to prove that cos⁡−135+cos⁡−11213+cos⁡−16365=π2\cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65} = \frac{\pi}{2}

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

    L.H.S. =cos⁡−135+cos⁡−11213+cos⁡−16365= \cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65}

    =cos⁡−1(35.1213−(1−3252)(1−122132))+cos⁡−16365= \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}

    =cos⁡−1(3665−45.513)+cos⁡−16365= \cos^{-1}\left(\frac{36}{65} - \frac{4}{5}.\frac{5}{13}\right) + \cos^{-1}\frac{63}{65}

    =cos⁡−1(3665−2065)+cos⁡−16365= \cos^{-1}\left(\frac{36}{65} - \frac{20}{65}\right) + \cos^{-1}\frac{63}{65}

    =cos⁡−11665+cos⁡−16365= \cos^{-1}\frac{16}{65} + \cos^{-1}\frac{63}{65}

    =cos⁡−1(1665.6364−(1−162652)(1−632652))= \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)

    =cos⁡−10=π2== \cos^{-1}0 = \frac{\pi}{2} = R.H.S.

  42. We have to prove that sin⁡−1x+sin⁡−1y=cos⁡−1(1−x21−y2−xy)\sin^{-1}x + \sin^{-1}y = \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right)

    L.H.S. =sin⁡−1x+sin⁡−1y=cos⁡−11−x2+cos⁡−11−y2= \sin^{-1}x + \sin^{-1}y = \cos^{-1}\sqrt{1 - x^2} + \cos^{-1}\sqrt{1 - y^2}

    =cos⁡1−(1−x21−y2−[1−(1−x2)][1−(1−y2)])= \cos^{1-}(\sqrt{1 - x^2}\sqrt{1 - y^2} - \sqrt{[1 - (1 - x^2)][1 - (1 - y^2)]})

    =cos⁡−1(1−x21−y2−xy)== \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right) = R.H.S.

  43. We have to prove that 4(sin⁡−1110+cos⁡−125)=π4\left(\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}}\right) =\pi

    or sin⁡−1110+cos⁡−125=π/4\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}} =\pi/4

    L.H.S. =sin⁡−1110<