\displaystyle \textbf{Question 1: }\text{Find the principal values of each of the following:}
\displaystyle \text{(i) }\cot^{-1}(-\sqrt3)\qquad \text{(ii) }\cot^{-1}(\sqrt3)
\displaystyle \text{(iii) }\cot^{-1}\left(-\frac{1}{\sqrt3}\right)\qquad \text{(iv) }\cot^{-1}\left(\tan\frac{3\pi}{4}\right)
\displaystyle \text{Answer:}
\displaystyle \text{For any }x\in\mathbb{R},\ \cot^{-1}x\text{ denotes an angle in }(0,\pi)\text{ whose cotangent is }x.
\displaystyle \text{(i) }\cot^{-1}(-\sqrt3)=\cot^{-1}\left(\cot\frac{5\pi}{6}\right)=\frac{5\pi}{6}.
\displaystyle \text{(ii) }\cot^{-1}(\sqrt3)=\cot^{-1}\left(\cot\frac{\pi}{6}\right)=\frac{\pi}{6}.
\displaystyle \text{(iii) }\cot^{-1}\left(-\frac{1}{\sqrt3}\right)=\cot^{-1}\left(\cot\frac{2\pi}{3}\right)=\frac{2\pi}{3}.
\displaystyle \text{(iv) }\tan\frac{3\pi}{4}=-1.
\displaystyle \therefore \cot^{-1}\left(\tan\frac{3\pi}{4}\right)=\cot^{-1}(-1)
\displaystyle =\cot^{-1}\left(\cot\frac{3\pi}{4}\right)=\frac{3\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the domain of }f(x)=\cot x+\cot^{-1}x.
\displaystyle \text{Answer:}
\displaystyle \text{The domain of }\cot x\text{ is }\mathbb{R}-\{n\pi:n\in\mathbb{Z}\}.
\displaystyle \text{The domain of }\cot^{-1}x\text{ is }\mathbb{R}.
\displaystyle \therefore \text{The required domain is }\left(\mathbb{R}-\{n\pi:n\in\mathbb{Z}\}\right)\cap\mathbb{R}.
\displaystyle =\mathbb{R}-\{n\pi:n\in\mathbb{Z}\}.
\displaystyle \therefore \text{The domain of }f(x)\text{ is }\mathbb{R}-\{n\pi:n\in\mathbb{Z}\}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Evaluate each of the following:}
\displaystyle \text{(i) }\cot^{-1}\left(\frac{1}{\sqrt3}\right)-\mathrm{cosec}^{-1}(-2)+\sec^{-1}\left(\frac{2}{\sqrt3}\right)
\displaystyle \text{(ii) }\cot^{-1}\left\{2\cos\left(\sin^{-1}\frac{\sqrt3}{2}\right)\right\}
\displaystyle \text{(iii) }\mathrm{cosec}^{-1}\left(-\frac{2}{\sqrt3}\right)+2\cot^{-1}(-1)
\displaystyle \text{(iv) }\tan^{-1}\left(-\frac{1}{\sqrt3}\right)+\cot^{-1}\left(-\frac{1}{\sqrt3}\right)+\tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right)
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\cot^{-1}\left(\frac{1}{\sqrt3}\right)-\mathrm{cosec}^{-1}(-2)+\sec^{-1}\left(\frac{2}{\sqrt3}\right)
\displaystyle =\cot^{-1}\left(\cot\frac{\pi}{3}\right)-\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{6}\right)\right)+\sec^{-1}\left(\sec\frac{\pi}{6}\right)
\displaystyle =\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)+\frac{\pi}{6}
\displaystyle =\frac{2\pi}{3}

\displaystyle \text{(ii) }\cot^{-1}\left\{2\cos\left(\sin^{-1}\frac{\sqrt3}{2}\right)\right\}
\displaystyle =\cot^{-1}\left\{2\cos\left(\sin^{-1}\left(\sin\frac{\pi}{3}\right)\right)\right\}
\displaystyle =\cot^{-1}\left(2\cos\frac{\pi}{3}\right)
\displaystyle =\cot^{-1}(1)
\displaystyle =\cot^{-1}\left(\cot\frac{\pi}{4}\right)
\displaystyle =\frac{\pi}{4}

\displaystyle \text{(iii) }\mathrm{cosec}^{-1}\left(-\frac{2}{\sqrt3}\right)+2\cot^{-1}(-1)
\displaystyle =\mathrm{cosec}^{-1}\left(\mathrm{cosec}\left(-\frac{\pi}{3}\right)\right)+2\cot^{-1}\left(\cot\frac{3\pi}{4}\right)
\displaystyle =-\frac{\pi}{3}+2\left(\frac{3\pi}{4}\right)
\displaystyle =-\frac{\pi}{3}+\frac{3\pi}{2}
\displaystyle =\frac{7\pi}{6}

\displaystyle \text{(iv) }\tan^{-1}\left(-\frac{1}{\sqrt3}\right)+\cot^{-1}\left(-\frac{1}{\sqrt3}\right)+\tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right)
\displaystyle =\tan^{-1}\left(\tan\left(-\frac{\pi}{6}\right)\right)+\cot^{-1}\left(\cot\frac{2\pi}{3}\right)+\tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right)
\displaystyle =-\frac{\pi}{6}+\frac{2\pi}{3}-\frac{\pi}{4}
\displaystyle =\frac{-2\pi+8\pi-3\pi}{12}
\displaystyle =\frac{3\pi}{12}
\displaystyle =\frac{\pi}{4}
\displaystyle \\

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