\displaystyle \textbf{Question 1: }\text{Find the principal values of each of the following:}
\displaystyle \text{(i) }\mathrm{cosec}^{-1}(-\sqrt2)\qquad \text{(ii) }\mathrm{cosec}^{-1}(-2)
\displaystyle \text{(iii) }\mathrm{cosec}^{-1}\left(\frac{2}{\sqrt3}\right)\qquad \text{(iv) }\mathrm{cosec}^{-1}\left(2\cos\frac{2\pi}{3}\right)
\displaystyle \text{Answer:}
\displaystyle \text{For }x\in(-\infty,-1]\cup[1,\infty),\ \mathrm{cosec}^{-1}x\text{ is an angle }\theta\in\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]
\displaystyle \text{such that }\mathrm{cosec}\,\theta=x.
\displaystyle \text{(i) }\mathrm{cosec}^{-1}(-\sqrt2)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{4}\right)\right)=-\frac{\pi}{4}.
\displaystyle \text{(ii) }\mathrm{cosec}^{-1}(-2)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{6}\right)\right)=-\frac{\pi}{6}.
\displaystyle \text{(iii) }\mathrm{cosec}^{-1}\left(\frac{2}{\sqrt3}\right)=\mathrm{cosec}^{-1}\left(\mathrm{cosec}\frac{\pi}{3}\right)=\frac{\pi}{3}.
\displaystyle \text{(iv) }2\cos\frac{2\pi}{3}=2\left(-\frac12\right)=-1.
\displaystyle \therefore \mathrm{cosec}^{-1}\left(2\cos\frac{2\pi}{3}\right)=\mathrm{cosec}^{-1}(-1)
\displaystyle =\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{2}\right)\right)=-\frac{\pi}{2}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the set of values of }\mathrm{cosec}^{-1}\left(\frac{\sqrt3}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \text{We know that }\mathrm{cosec}^{-1}x\text{ is defined for all }x\le-1\text{ or }x\ge1.
\displaystyle \text{But }\frac{\sqrt3}{2}<1.
\displaystyle \therefore \mathrm{cosec}^{-1}\left(\frac{\sqrt3}{2}\right)\text{ is not defined.}
\displaystyle \therefore \text{The set of values of }\mathrm{cosec}^{-1}\left(\frac{\sqrt3}{2}\right)\text{ is }\emptyset.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{For the principal values, evaluate the following:}
\displaystyle \text{(i) }\sin^{-1}\left(-\frac{\sqrt3}{2}\right)+\mathrm{cosec}^{-1}\left(-\frac{2}{\sqrt3}\right)
\displaystyle \text{(ii) }\sec^{-1}(\sqrt2)+2\mathrm{cosec}^{-1}(-\sqrt2)
\displaystyle \text{(iii) }\sin^{-1}\left[\cos\left\{2\mathrm{cosec}^{-1}(-2)\right\}\right]
\displaystyle \text{(iv) }\mathrm{cosec}^{-1}\left(2\tan\frac{11\pi}{6}\right)
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\sin^{-1}\left(-\frac{\sqrt3}{2}\right)+\mathrm{cosec}^{-1}\left(-\frac{2}{\sqrt3}\right)
\displaystyle =\sin^{-1}\left(\sin\left(-\frac{\pi}{3}\right)\right)+\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{3}\right)\right)
\displaystyle =-\frac{\pi}{3}-\frac{\pi}{3}
\displaystyle =-\frac{2\pi}{3}

\displaystyle \text{(ii) }\sec^{-1}(\sqrt2)+2\mathrm{cosec}^{-1}(-\sqrt2)
\displaystyle =\sec^{-1}\left(\sec\frac{\pi}{4}\right)+2\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{4}\right)\right)
\displaystyle =\frac{\pi}{4}-2\left(\frac{\pi}{4}\right)
\displaystyle =-\frac{\pi}{4}

\displaystyle \text{(iii) }\sin^{-1}\left[\cos\left\{2\mathrm{cosec}^{-1}(-2)\right\}\right]
\displaystyle =\sin^{-1}\left[\cos\left\{2\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{6}\right)\right)\right\}\right]
\displaystyle =\sin^{-1}\left(\cos\left(-\frac{\pi}{3}\right)\right)
\displaystyle =\sin^{-1}\left(\frac12\right)
\displaystyle =\sin^{-1}\left(\sin\frac{\pi}{6}\right)
\displaystyle =\frac{\pi}{6}

\displaystyle \text{(iv) }\mathrm{cosec}^{-1}\left(2\tan\frac{11\pi}{6}\right)
\displaystyle =\mathrm{cosec}^{-1}\left(2\cdot-\frac{1}{\sqrt3}\right)
\displaystyle =\mathrm{cosec}^{-1}\left(-\frac{2}{\sqrt3}\right)
\displaystyle =\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{3}\right)\right)
\displaystyle =-\frac{\pi}{3}
\displaystyle \\

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