Question 1: Find the principal values of each of the following:

\displaystyle \text{(i) }  \cot^{-1}  ( - \sqrt{3}) 

\displaystyle \text{(ii) } \cot^{-1}   (\sqrt{3})

\displaystyle \text{(iii) }  \cot^{-1} \Big( -\frac{1}{\sqrt{3}}  \Big)   

\displaystyle \text{(iv) } \cot^{-1} \Big( \tan \frac{3\pi}{4}  \Big)   

Answer:

\displaystyle \text{We know that for any } x \in R, \cot^{-1} x \text{ denotes an angle in } (0, \pi) \text{ whose cotangent is } x.

\displaystyle \text{(i) }  \cot^{-1}  ( - \sqrt{3})  \\ \\ = \Big( \text{An angle in } (0, \pi) \text{ whose cotangent is } -\sqrt{3}   = \cot \frac{5\pi}{6} \Big) = \frac{5\pi}{6}  

\displaystyle \text{(ii) } \cot^{-1}   (\sqrt{3}) \\ \\ = \Big( \text{An angle in } (0, \pi) \text{ whose cotangent is } \sqrt{3}   = \cot \frac{\pi}{6} \Big) = \frac{\pi}{6}  

\displaystyle \text{(iii) }  \cot^{-1} \Big( -\frac{1}{\sqrt{3}}  \Big)  \\ \\ = \Big( \text{An angle in } (0, \pi) \text{ whose cotangent is } -\frac{1}{\sqrt{3}}   = \cot \frac{2\pi}{3} \Big) = \frac{2\pi}{3}  

\displaystyle \text{(iv) } \cot^{-1} \Big( \tan \frac{3\pi}{4}  \Big) \\ \\ = \Big( \text{An angle in } (0, \pi) \text{ whose cotangent is } \tan \frac{3\pi}{4} = -1   = \cot \frac{3\pi}{4} \Big) = \frac{3\pi}{4}  

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Question 2: Find the domain of f(x) = \cot x + \cot^{-1} x 

Answer:

\displaystyle \text{The domain of } \cot x \text{ is } R - \{ n\pi , n \in Z \}

\displaystyle \text{The domain of } \cot^{-1} x \text{ is } R

\displaystyle \text{Therefore the domain of }  \cot x + \cot^{-1} x \\ \\ = R - \{ n\pi , n \in Z \} \cap R = R - \{ n\pi , n \in Z \}

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Question 3: Evaluate each of the following:

\displaystyle \text{(i) }  \cot^{-1}  \frac{1}{\sqrt{3}} - \mathrm{cosec}^{-1} ( -2) + \sec^{-1} \Big( \frac{2}{\sqrt{3}} \Big)

\displaystyle \text{(ii) } \cot^{-1} \Big\{  2 \cos \Big( \sin^{-1} \frac{\sqrt{3}}{2} \Big)   \Big\} 

\displaystyle \text{(iii) }  \mathrm{cosec}^{-1} \Big( -\frac{2}{\sqrt{3}} + 2 \cot^{-1} (-1) \Big)   

\displaystyle \text{(iv) } \tan^{-1} \Big( -\frac{1}{\sqrt{3}}  \Big) + \cot^{-1} \Big( -\frac{1}{\sqrt{3}}  \Big)  + \tan^{-1} \Big( \sin \Big( -\frac{\pi}{2} \Big) \Big)

Answer:

\displaystyle \text{(i) }  \cot^{-1}  \frac{1}{\sqrt{3}} - \mathrm{cosec}^{-1} ( -2) + \sec^{-1} \Big( \frac{2}{\sqrt{3}} \Big) \\ \\ \\ = \cot^{-1} \Big( \cot \frac{\pi}{3} \Big) - \mathrm{cosec}^{-1} \Big[ \mathrm{cosec} \Big( -\frac{\pi}{6} \Big)    \Big] + \sec^{-1} \Big( \sec \frac{\pi}{6} \Big) \\ \\ \\ = \frac{\pi}{3} + \frac{\pi}{6} + \frac{\pi}{6} \\ \\ \\ = \frac{2\pi}{3}

\displaystyle \text{(ii) } \cot^{-1} \Big\{  2 \cos \Big( \sin^{-1} \frac{\sqrt{3}}{2} \Big)   \Big\} \\ \\ \\ = \cot^{-1} \Big\{  2 \cos \Big( \sin^{-1} \Big( \sin \frac{\pi}{3} \Big)  \Big)   \Big\}   \\ \\ \\ = \cot^{-1} \Big( 2 \cos \frac{\pi}{3} \Big) \\ \\ \\ = \cot^{-1} \Big( 2 \times \frac{1}{2} \Big) \\ \\ \\ = \cot^{-1} ( 1) \\ \\ \\ = \cot^{-1} \Big( \tan \frac{\pi}{4} \Big) \\ \\ \\ =  \frac{\pi}{4}

\displaystyle \text{(iii) }  \mathrm{cosec}^{-1} \Big( -\frac{2}{\sqrt{3}} \Big) + 2 \cot^{-1} (-1)  \\ \\ \\ = \mathrm{cosec}^{-1} \Big( \mathrm{cosec} \Big( -\frac{\pi}{3} \Big)  \Big)+ 2 \cot^{-1} \Big[ \cot \Big( \frac{3\pi}{4} \Big) \Big]   \\ \\ \\ = -\frac{\pi}{3} + 2 \times \frac{3\pi}{4}\\ \\ \\ = -\frac{\pi}{3} + \frac{3\pi}{2} \\ \\ \\ =   \frac{7\pi}{6}

\displaystyle \text{(iv) } \tan^{-1} \Big( -\frac{1}{\sqrt{3}}  \Big) + \cot^{-1} \Big( -\frac{1}{\sqrt{3}}  \Big)  + \tan^{-1} \Big( \sin \Big( -\frac{\pi}{2} \Big) \Big) \\ \\ \\ =  \tan^{-1} \Big( \tan \Big(-\frac{\pi}{6} \Big)  \Big) + \cot^{-1} \Big( \cot \frac{\pi}{3}  \Big)  + \tan^{-1} (-1) \\ \\ \\ =  \tan^{-1} \Big( \tan \Big(-\frac{\pi}{6} \Big)  \Big) + \cot^{-1} \Big( \cot \frac{\pi}{3}  \Big)  + \tan^{-1} \Big(\tan \Big( - \frac{\pi}{4} \Big) \Big) \\ \\ \\ = - \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{4}  \\ \\ \\ =  - \frac{\pi}{12}

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