NCERT Solutions · Class 12 Mathematics Inverse Trigonometric Functions

The principal value branch of \(\displaystyle \cos^{-1}\) is \(\displaystyle [0,\pi]\), so the identity \(\displaystyle \cos^{-1}(\cos\theta)=\theta\) may be used only when \(\displaystyle \theta\in[0,\pi]\). Here \(\displaystyle \frac{13\pi}{6}>\pi\), so the angle must be reduced first.Using the \(\displaystyle 2\pi\)-periodicity of cosine, \[\frac{13\pi}{6}=2\pi+\frac{\pi}{6}\quad\Longrightarrow\quad \cos\frac{13\pi}{6}=\cos\left(2\pi+\frac{\pi}{6}\right)=\cos\frac{\pi}{6}.\]Now \(\displaystyle \frac{\pi}{6}\) does lie in \(\displaystyle [0,\pi]\), so the identity applies: \[\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\cos^{-1}\left(\cos\frac{\pi}{6}\right)=\frac{\pi}{6}.\]

The principal value branch of \(\displaystyle \tan^{-1}\) is \(\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), and \(\displaystyle \frac{7\pi}{6}\) is not in it, so \(\displaystyle \tan^{-1}\left(\tan\frac{7\pi}{6}\right)\neq\frac{7\pi}{6}\). Reduce the angle using the fact that tangent has period \(\displaystyle \pi\) (not \(\displaystyle 2\pi\)): \[\tan\frac{7\pi}{6}=\tan\left(\pi+\frac{\pi}{6}\right)=\tan\frac{\pi}{6}.\]Since \(\displaystyle \frac{\pi}{6}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), \[\tan^{-1}\left(\tan\frac{7\pi}{6}\right)=\tan^{-1}\left(\tan\frac{\pi}{6}\right)=\frac{\pi}{6}.\]

Let \(\displaystyle \theta=\sin^{-1}\frac{3}{5}\), so that \(\displaystyle \sin\theta=\frac{3}{5}\) with \(\displaystyle \theta\in\left(0,\frac{\pi}{2}\right)\).Since \(\displaystyle \theta\) is in the first quadrant, the cosine is taken positive: \[\cos\theta=\sqrt{1-\frac{9}{25}}=\frac{4}{5},\qquad \tan\theta=\frac{3}{4}.\]Apply the double angle formula \(\displaystyle \tan 2\theta=\dfrac{2\tan\theta}{1-\tan^{2}\theta}\): \[\tan 2\theta=\frac{2\cdot\dfrac{3}{4}}{1-\dfrac{9}{16}}=\frac{\dfrac{3}{2}}{\dfrac{7}{16}}=\frac{3}{2}\cdot\frac{16}{7}=\frac{24}{7}.\]Before concluding \(\displaystyle 2\theta=\tan^{-1}\frac{24}{7}\), check that \(\displaystyle 2\theta\) lies in the branch \(\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). Because \(\displaystyle \sin\theta=\frac{3}{5}=0.6<\frac{1}{\sqrt{2}}=\sin\frac{\pi}{4}\) and sine is increasing on \(\displaystyle \left(0,\frac{\pi}{2}\right)\), we get \(\displaystyle \theta<\frac{\pi}{4}\), hence \(\displaystyle 0<2\theta<\frac{\pi}{2}\). (This is exactly the step that fails when \(\displaystyle \sin^{-1}\) of a larger value is doubled.)Therefore \[2\sin^{-1}\frac{3}{5}=2\theta=\tan^{-1}\frac{24}{7}.\]

Let \(\displaystyle x=\sin^{-1}\frac{8}{17}\) and \(\displaystyle y=\sin^{-1}\frac{3}{5}\); both lie in \(\displaystyle \left(0,\frac{\pi}{2}\right)\), so all their cosines and tangents are positive.\[\sin x=\frac{8}{17}\Rightarrow\cos x=\sqrt{1-\frac{64}{289}}=\frac{15}{17}\Rightarrow\tan x=\frac{8}{15},\] \[\sin y=\frac{3}{5}\Rightarrow\cos y=\frac{4}{5}\Rightarrow\tan y=\frac{3}{4}.\]By the addition formula \(\displaystyle \tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\): \[\tan(x+y)=\frac{\dfrac{8}{15}+\dfrac{3}{4}}{1-\dfrac{8}{15}\cdot\dfrac{3}{4}}=\frac{\dfrac{32+45}{60}}{1-\dfrac{24}{60}}=\frac{\dfrac{77}{60}}{\dfrac{36}{60}}=\frac{77}{36}.\]Quadrant check: \(\displaystyle \tan x\tan y=\frac{24}{60}=\frac{2}{5}<1\) with both tangents positive, so \(\displaystyle x+y\) is an angle whose tangent is positive and which is less than \(\displaystyle \frac{\pi}{2}\); that is, \(\displaystyle x+y\in\left(0,\frac{\pi}{2}\right)\), the principal branch of \(\displaystyle \tan^{-1}\).Hence \[\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{77}{36}.\]

Let \(\displaystyle x=\cos^{-1}\frac{4}{5}\) and \(\displaystyle y=\cos^{-1}\frac{12}{13}\). Both cosines are positive, so \(\displaystyle x,y\in\left(0,\frac{\pi}{2}\right)\) and both sines are positive: \[\sin x=\sqrt{1-\frac{16}{25}}=\frac{3}{5},\qquad \sin y=\sqrt{1-\frac{144}{169}}=\frac{5}{13}.\]By the cosine addition formula \(\displaystyle \cos(x+y)=\cos x\cos y-\sin x\sin y\): \[\cos(x+y)=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}.\]The branch of \(\displaystyle \cos^{-1}\) is \(\displaystyle [0,\pi]\), and since \(\displaystyle 0<x<\frac{\pi}{2}\) and \(\displaystyle 0<y<\frac{\pi}{2}\) we have \(\displaystyle 0<x+y<\pi\). So \(\displaystyle x+y\) is the unique angle in \(\displaystyle [0,\pi]\) with cosine \(\displaystyle \frac{33}{65}\): \[\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=x+y=\cos^{-1}\frac{33}{65}.\]

Let \(\displaystyle x=\cos^{-1}\frac{12}{13}\) and \(\displaystyle y=\sin^{-1}\frac{3}{5}\), both in \(\displaystyle \left(0,\frac{\pi}{2}\right)\). Then \[\cos x=\frac{12}{13},\ \sin x=\frac{5}{13};\qquad \sin y=\frac{3}{5},\ \cos y=\frac{4}{5}.\]By the sine addition formula \(\displaystyle \sin(x+y)=\sin x\cos y+\cos x\sin y\): \[\sin(x+y)=\frac{5}{13}\cdot\frac{4}{5}+\frac{12}{13}\cdot\frac{3}{5}=\frac{20}{65}+\frac{36}{65}=\frac{56}{65}.\]The branch of \(\displaystyle \sin^{-1}\) is \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), so we must confirm \(\displaystyle x+y\) lies there before inverting. Compute \[\cos(x+y)=\cos x\cos y-\sin x\sin y=\frac{12}{13}\cdot\frac{4}{5}-\frac{5}{13}\cdot\frac{3}{5}=\frac{48-15}{65}=\frac{33}{65}>0,\] and \(\displaystyle x+y>0\); a positive cosine with \(\displaystyle 0<x+y<\pi\) forces \(\displaystyle 0<x+y<\frac{\pi}{2}\).Therefore \[\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}=x+y=\sin^{-1}\frac{56}{65}.\]

Work on the right-hand side. Let \(\displaystyle x=\sin^{-1}\frac{5}{13}\) and \(\displaystyle y=\cos^{-1}\frac{3}{5}\); both lie in \(\displaystyle \left(0,\frac{\pi}{2}\right)\), so all ratios below are positive.\[\sin x=\frac{5}{13}\Rightarrow\cos x=\frac{12}{13}\Rightarrow\tan x=\frac{5}{12},\] \[\cos y=\frac{3}{5}\Rightarrow\sin y=\frac{4}{5}\Rightarrow\tan y=\frac{4}{3}.\]By the addition formula for tangent, \[\tan(x+y)=\frac{\dfrac{5}{12}+\dfrac{4}{3}}{1-\dfrac{5}{12}\cdot\dfrac{4}{3}}=\frac{\dfrac{5+16}{12}}{1-\dfrac{20}{36}}=\frac{\dfrac{21}{12}}{\dfrac{16}{36}}=\frac{21}{12}\cdot\frac{36}{16}=\frac{63}{16}.\]Since \(\displaystyle \tan x\tan y=\frac{20}{36}=\frac{5}{9}<1\) with both tangents positive, \(\displaystyle x+y\in\left(0,\frac{\pi}{2}\right)\), which is inside the principal branch of \(\displaystyle \tan^{-1}\). Hence \(\displaystyle x+y=\tan^{-1}\frac{63}{16}\), i.e. \[\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}.\]

Substitute \(\displaystyle \sqrt{x}=\tan\theta\), i.e. \(\displaystyle \theta=\tan^{-1}\sqrt{x}\), which is the left-hand side.Range of the substitution: for \(\displaystyle x\in[0,1]\) we have \(\displaystyle \sqrt{x}\in[0,1]\), so \(\displaystyle \theta\in\left[0,\frac{\pi}{4}\right]\) and therefore \(\displaystyle 2\theta\in\left[0,\frac{\pi}{2}\right]\). This is the step the whole proof rests on.Since \(\displaystyle x=\tan^{2}\theta\), \[\frac{1-x}{1+x}=\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos 2\theta,\] using the standard identity \(\displaystyle \cos 2\theta=\dfrac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\).Hence \[\cos^{-1}\frac{1-x}{1+x}=\cos^{-1}(\cos 2\theta)=2\theta,\] the last equality being legitimate because \(\displaystyle 2\theta\in\left[0,\frac{\pi}{2}\right]\subseteq[0,\pi]\), the principal branch of \(\displaystyle \cos^{-1}\).Therefore \[\frac{1}{2}\cos^{-1}\frac{1-x}{1+x}=\theta=\tan^{-1}\sqrt{x},\qquad x\in[0,1].\]

Write both surds using the half-angle identities \[1+\sin x=\cos^{2}\frac{x}{2}+\sin^{2}\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}=\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^{2},\] \[1-\sin x=\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}.\]Taking square roots requires the sign of each bracket, and this is where the given domain is used. For \(\displaystyle x\in\left(0,\frac{\pi}{4}\right)\) we have \(\displaystyle \frac{x}{2}\in\left(0,\frac{\pi}{8}\right)\), so \(\displaystyle \cos\frac{x}{2}>\sin\frac{x}{2}>0\) and both brackets are positive: \[\sqrt{1+\sin x}=\cos\frac{x}{2}+\sin\frac{x}{2},\qquad \sqrt{1-\sin x}=\cos\frac{x}{2}-\sin\frac{x}{2}.\]Substituting, \[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{2\cos\dfrac{x}{2}}{2\sin\dfrac{x}{2}}=\cot\frac{x}{2}.\]Since \(\displaystyle \frac{x}{2}\in\left(0,\frac{\pi}{8}\right)\subset(0,\pi)\), the principal branch of \(\displaystyle \cot^{-1}\), we may cancel the functions: \[\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}.\]

Follow the hint: put \(\displaystyle x=\cos 2\theta\), so that \(\displaystyle 2\theta=\cos^{-1}x\) and \(\displaystyle \theta=\frac{1}{2}\cos^{-1}x\).Fix the range of \(\displaystyle \theta\) first. Since \(\displaystyle \cos^{-1}\) takes values in \(\displaystyle [0,\pi]\) and \(\displaystyle x\in\left[-\frac{1}{\sqrt{2}},1\right]\), \[2\theta=\cos^{-1}x\in\left[0,\frac{3\pi}{4}\right]\quad\Longrightarrow\quad \theta\in\left[0,\frac{3\pi}{8}\right],\] so \(\displaystyle \cos\theta>0\) and \(\displaystyle \sin\theta\geq 0\).Now use \(\displaystyle 1+\cos 2\theta=2\cos^{2}\theta\) and \(\displaystyle 1-\cos 2\theta=2\sin^{2}\theta\): \[\sqrt{1+x}=\sqrt{2}\,\cos\theta,\qquad \sqrt{1-x}=\sqrt{2}\,\sin\theta,\] the positive roots being correct precisely because of the range just established.Therefore \[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\frac{1-\tan\theta}{1+\tan\theta}=\tan\left(\frac{\pi}{4}-\theta\right),\] dividing numerator and denominator by \(\displaystyle \cos\theta\) and using \(\displaystyle \tan\frac{\pi}{4}=1\).Finally \(\displaystyle \frac{\pi}{4}-\theta\in\left[\frac{\pi}{4}-\frac{3\pi}{8},\frac{\pi}{4}\right]=\left[-\frac{\pi}{8},\frac{\pi}{4}\right]\), which lies inside \(\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), so \[\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x.\]