\displaystyle \textbf{Question 1: }\text{Find the principal values for each of the following:}
\displaystyle \text{(i) }\sec^{-1}(-\sqrt2)\qquad \text{(ii) }\sec^{-1}(2)
\displaystyle \text{(iii) }\sec^{-1}\left(2\sin\frac{3\pi}{4}\right)\qquad \text{(iv) }\sec^{-1}\left(2\tan\frac{3\pi}{4}\right)
\displaystyle \text{Answer:}
\displaystyle \text{For any }x\in(-\infty,-1]\cup[1,\infty),\ \sec^{-1}x\text{ is an angle }\theta\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]\text{ such that }\sec\theta=x.

\displaystyle \text{(i) }\sec^{-1}(-\sqrt2)=\sec^{-1}\left(\sec\frac{3\pi}{4}\right)=\frac{3\pi}{4}.

\displaystyle \text{(ii) }\sec^{-1}(2)=\sec^{-1}\left(\sec\frac{\pi}{3}\right)=\frac{\pi}{3}.

\displaystyle \text{(iii) }\sec^{-1}\left(2\sin\frac{3\pi}{4}\right)=\sec^{-1}(\sqrt2)
\displaystyle =\sec^{-1}\left(\sec\frac{\pi}{4}\right)=\frac{\pi}{4}.

\displaystyle \text{(iv) }\sec^{-1}\left(2\tan\frac{3\pi}{4}\right)=\sec^{-1}(-2)
\displaystyle =\sec^{-1}\left(\sec\frac{2\pi}{3}\right)=\frac{2\pi}{3}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the principal values and evaluate the following:}
\displaystyle \text{(i) }\tan^{-1}\sqrt3-\sec^{-1}(-2)\hfill\text{[CBSE 2012]}
\displaystyle \text{(ii) }\sin^{-1}\left(-\frac{\sqrt3}{2}\right)-2\sec^{-1}\left(2\tan\frac{\pi}{6}\right)
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\tan^{-1}\sqrt3-\sec^{-1}(-2)
\displaystyle =\tan^{-1}\left(\tan\frac{\pi}{3}\right)-\sec^{-1}\left(\sec\frac{2\pi}{3}\right)
\displaystyle =\frac{\pi}{3}-\frac{2\pi}{3}
\displaystyle =-\frac{\pi}{3}

\displaystyle \text{(ii) }\sin^{-1}\left(-\frac{\sqrt3}{2}\right)-2\sec^{-1}\left(2\tan\frac{\pi}{6}\right)
\displaystyle =\sin^{-1}\left(\sin\left(-\frac{\pi}{3}\right)\right)-2\sec^{-1}\left(\frac{2}{\sqrt3}\right)
\displaystyle =-\frac{\pi}{3}-2\sec^{-1}\left(\sec\frac{\pi}{6}\right)
\displaystyle =-\frac{\pi}{3}-2\left(\frac{\pi}{6}\right)
\displaystyle =-\frac{2\pi}{3}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the domain of:}
\displaystyle \text{(i) }\sec^{-1}(3x-1)
\displaystyle \text{(ii) }\sec^{-1}x-\tan^{-1}x
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\sec^{-1}(3x-1)
\displaystyle \text{The domain of }\sec^{-1}x\text{ is }(-\infty,-1]\cup[1,\infty).
\displaystyle \therefore 3x-1\le-1\text{ or }3x-1\ge1.
\displaystyle \therefore 3x\le0\text{ or }3x\ge2.
\displaystyle \therefore x\le0\text{ or }x\ge\frac23.
\displaystyle \therefore \text{The domain is }(-\infty,0]\cup\left[\frac23,\infty\right).

\displaystyle \text{(ii) }\sec^{-1}x-\tan^{-1}x
\displaystyle \text{The domain of }\sec^{-1}x\text{ is }(-\infty,-1]\cup[1,\infty).
\displaystyle \text{The domain of }\tan^{-1}x\text{ is }\mathbb{R}.
\displaystyle \therefore \text{The required domain is }\left((-\infty,-1]\cup[1,\infty)\right)\cap\mathbb{R}.
\displaystyle =(-\infty,-1]\cup[1,\infty).
\displaystyle \therefore \text{The domain is }(-\infty,-1]\cup[1,\infty).
\displaystyle \\

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