NCERT Solutions · Class 12 Mathematics Inverse Trigonometric Functions

Evaluate each inverse function separately, each in its own principal value branch, and only then add.\(\displaystyle \tan^{-1}(1)\): need \(\displaystyle y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) with \(\displaystyle \tan y=1\), so \(\displaystyle \tan^{-1}(1)=\frac{\pi}{4}\).\(\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)\): need \(\displaystyle y\in[0,\pi]\) with \(\displaystyle \cos y=-\frac{1}{2}\); since \(\displaystyle \cos\left(\pi-\frac{\pi}{3}\right)=-\frac{1}{2}\), this is \(\displaystyle \frac{2\pi}{3}\).\(\displaystyle \sin^{-1}\left(-\frac{1}{2}\right)\): need \(\displaystyle y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) with \(\displaystyle \sin y=-\frac{1}{2}\), which is \(\displaystyle -\frac{\pi}{6}\). Note the two negative arguments give answers of opposite sign — \(\displaystyle \cos^{-1}\) never returns a negative angle, \(\displaystyle \sin^{-1}\) does.Adding, with denominator $\displaystyle 12$, \[\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}=\frac{3\pi+8\pi-2\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}{4}.\]Value: \(\displaystyle \dfrac{3\pi}{4}\).

Evaluate each inverse function in its principal value branch first.\(\displaystyle \cos^{-1}\frac{1}{2}\): the angle \(\displaystyle y\in[0,\pi]\) with \(\displaystyle \cos y=\frac{1}{2}\) is \(\displaystyle y=\frac{\pi}{3}\).\(\displaystyle \sin^{-1}\frac{1}{2}\): the angle \(\displaystyle y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) with \(\displaystyle \sin y=\frac{1}{2}\) is \(\displaystyle y=\frac{\pi}{6}\).Therefore \[\cos^{-1}\frac{1}{2}+2\sin^{-1}\frac{1}{2}=\frac{\pi}{3}+2\cdot\frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}.\]Value: \(\displaystyle \dfrac{2\pi}{3}\).

By definition, \(\displaystyle \sin^{-1}\) is the inverse of the restriction of \(\displaystyle \sin\) to the principal value branch \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), on which sine is one-one and onto \(\displaystyle [-1,1]\). So if \(\displaystyle \sin^{-1}x=y\), then by definition \[y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].\]The interval is closed, not open: at \(\displaystyle x=1\) we get \(\displaystyle y=\frac{\pi}{2}\) and at \(\displaystyle x=-1\) we get \(\displaystyle y=-\frac{\pi}{2}\), both of which are attained. Hence (D) is wrong. Options (A) and (C) describe the range of \(\displaystyle \cos^{-1}\) and of \(\displaystyle \cot^{-1}\), not of \(\displaystyle \sin^{-1}\).Correct option: (B) \(\displaystyle -\frac{\pi}{2}\le y\le\frac{\pi}{2}\).

Take the two terms separately, each in its own principal value branch.\(\displaystyle \tan^{-1}\sqrt{3}\): need \(\displaystyle y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) with \(\displaystyle \tan y=\sqrt{3}\), so \(\displaystyle \tan^{-1}\sqrt{3}=\frac{\pi}{3}\).\(\displaystyle \sec^{-1}(-2)\): the branch is \(\displaystyle [0,\pi]-\left\{\frac{\pi}{2}\right\}\), and \[\sec y=-2\iff \cos y=-\frac{1}{2}.\] Since \(\displaystyle \cos\left(\pi-\frac{\pi}{3}\right)=-\frac{1}{2}\), we get \(\displaystyle \sec^{-1}(-2)=\frac{2\pi}{3}\). (Writing \(\displaystyle -\frac{\pi}{3}\) here is the usual slip; \(\displaystyle \sec^{-1}\) never returns a negative angle.)Hence \[\tan^{-1}\sqrt{3}-\sec^{-1}(-2)=\frac{\pi}{3}-\frac{2\pi}{3}=-\frac{\pi}{3}.\]Correct option: (B) \(\displaystyle -\frac{\pi}{3}\).