Question 1: Find the principal values for each one of the following: $\displaystyle \text{i) } \sec^{-1} ( - \sqrt{2})$ $\displaystyle \text{ii) } \sec^{-1} ( 2 )$ $\displaystyle \text{iii) } \sec^{-1} \Bigg(2 \sin \frac{3\pi}{4} \Bigg)$ $\displaystyle \text{iv) } \sec^{-1} \Bigg( 2 \tan \frac{3\pi}{4} \Bigg)$ $\displaystyle \text{For any } x \in ( -\infty, -1] \cup [1, \infty), \sec^{-1} x \text{ is an angle } \theta \in \Big[0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2}, \pi \Big] \\ \\ \text{ whose secant is } x \text{ i.e. } \sec \theta = x. \text{ Therefore, }$ $\displaystyle \text{i) } \sec^{-1} ( - \sqrt{2}) \\ \\ \\ = \Big( \text{An angle } \theta \in \Big[0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2}, \pi \Big] \text{ such that } \sec \theta = -\sqrt{2} = \sec \Big( \frac{3\pi}{4} \Big) \Big) = \frac{3\pi}{4}$ $\displaystyle \text{ii) } \sec^{-1} ( 2 ) \\ \\ \\ = \Big( \text{An angle } \theta \in \Big[0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2}, \pi \Big] \text{ such that } \sec \theta = 2 = \sec \Big( \frac{\pi}{3} \Big) \Big) = \frac{\pi}{3}$ $\displaystyle \text{iii) } \sec^{-1} \Bigg(2 \sin \frac{3\pi}{4} \Bigg) \\ \\ \\ = \Bigg( \text{An angle } \theta \in \Big[0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2}, \pi \Big] \text{ such that } \sec \theta = 2 \sin \frac{3\pi}{4} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} = \sec \Big( \frac{\pi}{4} \Big) \Bigg) = \frac{\pi}{4}$ $\displaystyle \text{iv) } \sec^{-1} \Bigg( 2 \tan \frac{3\pi}{4} \Bigg) \\ \\ \\ = \Bigg( \text{An angle } \theta \in \Big[0, \frac{\pi}{2} \Big) \cup \Big( \frac{\pi}{2}, \pi \Big] \text{ such that } \sec \theta = 2 \tan \frac{3\pi}{4} = 2 \times (-1) = -2 = \sec \Big( \frac{2\pi}{3} \Big) \Bigg) = \frac{2\pi}{3}$ $\\$

Question 2: Find the principal values, evaluate the following: $\displaystyle \text{i) } \tan^{-1} \sqrt{3} - \sec^{-1} (-2) \hspace{2.0cm} \textbf{ CBSE 2012}$ $\displaystyle \text{ii) } \sin^{-1} \Bigg( - \frac{\sqrt{3}}{2} \Bigg) - 2 \sec^{-1} \Bigg( 2 \tan \frac{\pi}{6} \Bigg)$ $\displaystyle \text{i) } \tan^{-1} \sqrt{3} - \sec^{-1} (-2) \\ \\ \\ = \tan^{-1} \Big( \tan \frac{\pi}{3} \Big) - \sec^{-1} \Big(\sec \frac{2\pi}{3} \Big) \\ \\ \\ = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$ $\displaystyle \text{ii) } \sin^{-1} \Bigg( - \frac{\sqrt{3}}{2} \Bigg) - 2 \sec^{-1} \Bigg( 2 \tan \frac{\pi}{6} \Bigg) \\ \\ \\ = -\sin^{-1} \Bigg( \frac{\sqrt{3}}{2} \Bigg) - 2 \sec^{-1} \Bigg( 2 \times \frac{1}{\sqrt{3}} \Bigg) \\ \\ \\ = -\sin^{-1} \Bigg( \frac{\sqrt{3}}{2} \Bigg) - 2 \sec^{-1} \Bigg( \frac{2}{\sqrt{3}} \Bigg) \\ \\ \\ = -\sin^{-1} \Bigg( \sin \frac{\pi}{3} \Bigg) - 2 \sec^{-1} \Bigg( \sec\frac{\pi}{6} \Bigg) \\ \\ \\ = - \frac{\pi}{3} - \frac{\pi}{3} = - \frac{2\pi}{3}$ $\\$

Question 3: Find the domain of: $\displaystyle \text{i) } \sec^{-1} (3x-1)$ $\displaystyle \text{ii) } \sec^{-1} x - \tan^{-1} x$ $\displaystyle \text{i) } \sec^{-1} (3x-1)$ $\text{Domain of } \sec^{-1} \text{ is } (-\infty, -1] \cup [1, \infty).$ $\displaystyle \therefore \text{The Domain of } \sec^{-1} (3x-1) \text{ lies in the interval } (-\infty, -1] \cup [1, \infty)$ $\displaystyle \Rightarrow -\infty \leq 3x-1 \leq -1 \text{ and } 1 \leq 3x-1 \leq \infty$ $\displaystyle \Rightarrow -\infty \leq 3x \leq 0 \text{ and } 2 \leq 3x \leq \infty$ $\displaystyle \Rightarrow -\infty \leq x \leq 0 \text{ and } \frac{2}{3} \leq x \leq \infty$ $\displaystyle \text{ii) } \sec^{-1} x - \tan^{-1} x$ $\text{Domain of } \sec^{-1} \text{ is } (-\infty, -1] \cup [1, \infty). \hspace{1.0cm} \ldots (i)$ $\text{Domain of } \tan^{-1} \text{ is } R \hspace{1.0cm} \ldots (ii)$

Union of (i) and (ii)  will be the domain of the given function. $(-\infty, -1] \cup [1, \infty) \cup R \Rightarrow (-\infty, -1] \cup [1, \infty).$

Therefore the domain of the given function is $(-\infty, -1] \cup [1, \infty)$