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

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} (-\sqrt{2})$

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} (-2)$

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} \Bigg( \frac{2}{\sqrt{3}} \Bigg)$

$\displaystyle \text{(i) } \mathrm{cosec}^{-1} \Bigg( 2 \cos \frac{2\pi}{3} \Bigg)$

$\displaystyle \text{For } x \in ( -\infty, - 1] \ \cup \ [ 1 , \infty ) , \mathrm{cosec}^{-1} \text{ is an angle } \theta \in \Big[ -\frac{\pi}{2}, 0 \Big) \cup \Big( 0, \frac{\pi}{2} \Big] \\ \\ \text{ such that } \mathrm{cosec} \theta = x .$

$\text{(i) } \mathrm{cosec}^{-1} (-\sqrt{2}) \\ \\ = \Big( \text{An angle } \theta \in \Big[ -\frac{\pi}{2}, 0 \Big) \cup \Big( 0, \frac{\pi}{2} \Big] \text{ such that } \mathrm{cosec} \theta = (-\sqrt{2}) = \mathrm{cosec} \Big( - \frac{\pi}{4} \Big) \Big) \\ \\ { \displaystyle = - \frac{\pi}{4} }$

$\text{(ii) } \mathrm{cosec}^{-1} (-2) \\ \\ = \Big( \text{An angle } \theta \in \Big[ -\frac{\pi}{2}, 0 \Big) \cup \Big( 0, \frac{\pi}{2} \Big] \text{ such that } \mathrm{cosec} \theta = (-2) = \mathrm{cosec} \Big( - \frac{\pi}{6} \Big) \Big) \\ \\ { \displaystyle = - \frac{\pi}{6} }$

$\text{(iii) } \mathrm{cosec}^{-1} \Bigg( \frac{2}{\sqrt{3}} \Bigg) \\ \\ = \Big( \text{An angle } \theta \in \Big[ -\frac{\pi}{2}, 0 \Big) \cup \Big( 0, \frac{\pi}{2} \Big] \text{ such that } \mathrm{cosec} \theta = \Bigg( \frac{2}{\sqrt{3}} \Bigg) = \mathrm{cosec} \Big( \frac{\pi}{3} \Big) \Big) \\ \\ { \displaystyle = \frac{\pi}{3} }$

$\text{(iv) } \mathrm{cosec}^{-1} \Big( 2 \cos \frac{2\pi}{3} \Big) \\ \\ = \Big( \text{An angle } \theta \in \Big[ -\frac{\pi}{2}, 0 \Big) \cup \Big( 0, \frac{\pi}{2} \Big] \text{ such that } \mathrm{cosec} \theta = \Big( 2 \cos \frac{2\pi}{3} \Big) = \mathrm{cosec} \Big( -\frac{\pi}{3} \Big) \Big) \\ \\ { \displaystyle = -\frac{\pi}{2} }$

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$\displaystyle \text{Question 2: Find the set of values of: } \mathrm{cosec}^{-1} \Bigg( \frac{\sqrt{3}}{2} \Bigg)$

$\displaystyle \text{We know that } \mathrm{cosec}^{-1} x \text{ is defined for all } x \leq -1 \text{ or } x \geq 1.$

$\displaystyle \text{But } \frac{\sqrt{3}}{2} < 1$

$\displaystyle \text{So, } \mathrm{cosec}^{-1} \Bigg( \frac{\sqrt{3}}{2} \Bigg) \text{ is not meaningful. Hence, the set of values of } \mathrm{cosec}^{-1} \Bigg( \frac{\sqrt{3}}{2} \Bigg) \\ \\ \text{ is the null } \ \ set \ \ \emptyset$

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Question 3: For the principal values, evaluate the following:

$\displaystyle \text{(i) } \sin^{-1} \Bigg(- \frac{\sqrt{3}}{2} \Bigg) + \mathrm{cosec}^{-1} \Bigg( -\frac{2}{\sqrt{3}}\Bigg)$

$\displaystyle \text{(ii) } \sec^{-1} (\sqrt{2}) + 2\mathrm{cosec}^{-1} (-\sqrt{2})$

$\displaystyle \text{(iii) } \sin^{-1} \Bigg[ \cos \Bigg\{ 2 \mathrm{cosec}^{-1} ( - 2) \Bigg\} \Bigg]$

$\displaystyle \text{(iv) } \mathrm{cosec}^{-1} \Bigg( 2 \tan \frac{11\pi}{6} \Bigg)$

$\displaystyle \text{(i) } \sin^{-1} \Bigg(- \frac{\sqrt{3}}{2} \Bigg) + \mathrm{cosec}^{-1} \Bigg( -\frac{2}{\sqrt{3}}\Bigg) \\ \\ \\ = -\sin^{-1} \Bigg( \frac{\sqrt{3}}{2} \Bigg) + \mathrm{cosec}^{-1} \Bigg( -\frac{2}{\sqrt{3}}\Bigg) \\ \\ \\ = -\sin^{-1} \Bigg( \sin \frac{\pi}{3} \Bigg) + \mathrm{cosec}^{-1} \Bigg( \mathrm{cosec} \Big(-\frac{\pi}{3} \Big)\Bigg) \\ \\ \\ = - \frac{\pi}{3} - \frac{\pi}{3} \\ \\ \\ = - \frac{2\pi}{3}$

$\displaystyle \text{(ii) } \sec^{-1} (\sqrt{2}) + 2\mathrm{cosec}^{-1} (-\sqrt{2}) \\ \\ \\ = \sec^{-1} ( \sec\frac{\pi}{4} ) + 2 \mathrm{cosec}^{-1} \Bigg( \mathrm{cosec} \Big(-\frac{\pi}{4} \Big)\Bigg) \\ \\ \\ =\frac{\pi}{4} - 2 \times \frac{\pi}{4} \\ \\ \\ =\frac{\pi}{4} - \frac{\pi}{2} \\ \\ \\ = - \frac{\pi}{4}$

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

$\displaystyle \text{(iv) } \mathrm{cosec}^{-1} \Bigg( 2 \tan \frac{11\pi}{6} \Bigg) \\ \\ \\ = \mathrm{cosec}^{-1} \Bigg( 2 \times \frac{-1}{\sqrt3} \Bigg) \\ \\ \\ = \mathrm{cosec}^{-1} \Bigg( \frac{-2}{\sqrt3} \Bigg) \\ \\ \\ = \mathrm{cosec}^{-1} \Bigg( \mathrm{cosec} \frac{-\pi}{3}\Bigg) \\ \\ \\ = \frac{-\pi}{3}$