Question 1: Evaluate each of the following:

$\displaystyle \text{(i) } \sin \Bigg( \sin^{-1} \frac{7}{25} \Bigg) \hspace{2.0cm} \text{(ii) } \sin \Bigg( \cos^{-1} \frac{5}{13} \Bigg)$

$\displaystyle \text{(iii) } \sin \Bigg( \tan^{-1} \frac{24}{7} \Bigg) \hspace{2.0cm} \text{(iv) } \sin \Bigg( \sec^{-1} \frac{17}{8} \Bigg)$

$\displaystyle \text{(v) } \mathrm{cosec} \Bigg( \cos^{-1} \frac{3}{5} \Bigg) \hspace{2.0cm} \text{(vi) } \sec \Bigg( \sin^{-1} \frac{12}{13} \Bigg)$

$\displaystyle \text{(vii) } \tan \Bigg( \cos^{-1} \frac{8}{17} \Bigg) \hspace{2.0cm} \text{(viii) } \cot \Bigg( \cos^{-1} \frac{3}{5} \Bigg)$

$\displaystyle \text{(ix) } \cos \Bigg( \tan^{-1} \frac{24}{7} \Bigg)$

$\displaystyle \text{(i) } \sin \Bigg( \sin^{-1} \frac{7}{25} \Bigg) = \frac{7}{25}$

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$\displaystyle \text{(ii) } \sin \Bigg( \cos^{-1} \frac{5}{13} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \sqrt{1 - \Big( \frac{5}{13} \Big)^2} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \sqrt{1 - \frac{25}{169} } \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \sqrt{\frac{144}{169} } \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \frac{12}{13} \Bigg)$

$\displaystyle = \frac{12}{13}$

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$\displaystyle \text{(iii) } \sin \Bigg( \tan^{-1} \frac{24}{7} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \frac{\frac{24}{7}}{\sqrt{1 + \Big( \frac{24}{7} \Big)^2}} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \frac{\frac{24}{7}}{\sqrt{1 + \frac{576}{49} }} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \frac{\frac{24}{7}}{\sqrt{\frac{625}{49} } } \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \frac{\frac{24}{7}}{\frac{25}{7}} \Bigg)$

$\displaystyle = \frac{24}{25}$

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$\displaystyle \text{(iv) } \sin \Bigg( \sec^{-1} \frac{17}{8} \Bigg)$

$\displaystyle = \sin \Bigg( \cos^{-1} \frac{8}{17} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \sqrt{1 - \Big( \frac{8}{17} \Big)^2} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \sqrt{1 - \frac{64}{289} } \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \sqrt{\frac{225}{289} } \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \frac{15}{17} \Bigg)$

$\displaystyle = \frac{15}{17}$

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$\displaystyle \text{(v) } \mathrm{cosec} \Bigg( \cos^{-1} \frac{3}{5} \Bigg)$

$\displaystyle = \mathrm{cosec} \Bigg( \sin^{-1} \sqrt{1 - \Big( \frac{3}{5} \Big)^2} \Bigg)$

$\displaystyle = \mathrm{cosec} \Bigg( \sin^{-1} \sqrt{1 - \frac{9}{25} } \Bigg)$

$\displaystyle = \mathrm{cosec} \Bigg( \sin^{-1} \sqrt{\frac{16}{25} } \Bigg)$

$\displaystyle = \mathrm{cosec} \Bigg( \sin^{-1} \frac{4}{5} \Bigg)$

$\displaystyle = \mathrm{cosec} \Bigg( \mathrm{cosec}^{-1} \frac{5}{4} \Bigg)$

$\displaystyle = \frac{5}{4}$

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$\displaystyle \text{(vi) } \sec \Bigg( \sin^{-1} \frac{12}{13} \Bigg)$

$\displaystyle = \sec \Bigg( \cos^{-1} \sqrt{1 - \Big( \frac{12}{13} \Big)^2} \Bigg)$

$\displaystyle = \sec \Bigg( \cos^{-1} \sqrt{1 - \frac{144}{169} } \Bigg)$

$\displaystyle = \sec \Bigg( \cos^{-1} \sqrt{\frac{25}{169} } \Bigg)$

$\displaystyle = \sec \Bigg( \cos^{-1} \frac{5}{13} \Bigg)$

$\displaystyle = \sec \Bigg( \sec^{-1} \frac{13}{5} \Bigg)$

$\displaystyle = \frac{13}{5}$

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$\displaystyle \text{(vii) } \tan \Bigg( \cos^{-1} \frac{8}{17} \Bigg)$

$\displaystyle = \tan \Bigg( \tan^{-1} \frac{\sqrt{1 - \Big( \frac{8}{17} \Big)^2}}{\frac{8}{17}} \Bigg) = \tan \Bigg( \tan^{-1} \frac{\frac{15}{17}}{\frac{8}{17}} \Bigg) = \frac{15}{8}$

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$\displaystyle \text{(viii) } \cot \Bigg( \cos^{-1} \frac{3}{5} \Bigg)$

$\displaystyle = \cot \Bigg( \tan^{-1} \frac{\sqrt{1 - \Big( \frac{3}{5} \Big)^2}}{\frac{3}{5}} \Bigg) = \cot \Bigg( \tan^{-1} \frac{\frac{4}{5}}{\frac{3}{5}} \Bigg) = \cot \Bigg( \cot^{-1} \frac{3}{4} \Bigg) = \frac{3}{4}$

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$\displaystyle \text{(ix) } \cos \Bigg( \tan^{-1} \frac{24}{7} \Bigg)$

$\displaystyle = \cos \Bigg( \cos^{-1} \frac{1}{\sqrt{1 + \Big( \frac{24}{7} \Big)^2}} \Bigg)$

$\displaystyle = \cos \Bigg( \cos^{-1} \frac{1}{\sqrt{1 + \frac{576}{49} }} \Bigg)$

$\displaystyle = \cos \Bigg( \cos^{-1} \frac{1}{\sqrt{\frac{625}{49} } } \Bigg)$

$\displaystyle = \cos \Bigg( \cos^{-1} \frac{1}{\frac{25}{7}} \Bigg)$

$\displaystyle = \frac{7}{25}$

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Question 2: Prove the following results:

$\displaystyle \text{(i) } \tan \Bigg( \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{2}{3} \Bigg) = \frac{17}{16}$

$\displaystyle \text{(i) } \cos \Bigg( \sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2} \Bigg) = \frac{5}{5\sqrt{13}} \hspace{1.0cm} \textbf{ [CBSE 2012 ]}$

$\displaystyle \text{(i) } \tan \Bigg( \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5} \Bigg) = \frac{63}{16}$

$\displaystyle \text{(i) } \sin \Bigg( \cos^{-1} \frac{3}{5} + \sin^{-1} \frac{5}{13} \Bigg) = \frac{63}{65}$

$\displaystyle \text{(i) LHS } = \tan \Bigg( \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{2}{3} \Bigg)$

$\displaystyle = \frac{\tan \Big( \cos^{-1} \frac{4}{5} \Big) + \tan \Big( \tan^{-1} \frac{2}{3} \Big) }{1 - \tan \Big( \cos^{-1} \frac{4}{5} \Big) \tan \Big( \tan^{-1} \frac{2}{3} \Big) }$

$\displaystyle = \frac{\tan \Big( \tan^{-1} \frac{3}{4} \Big) + \tan \Big( \tan^{-1} \frac{2}{3} \Big) }{1 - \tan \Big( \tan^{-1} \frac{3}{4} \Big) \tan \Big( \tan^{-1} \frac{2}{3} \Big) }$

$\displaystyle = \frac{\frac{3}{4}+ \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}} = \frac{17}{6} = \text{ RHS. Hence, proved. }$

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$\displaystyle \text{(ii) LHS } = \cos \Bigg( \sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2} \Bigg)$

$\displaystyle = \cos \Bigg( \sin^{-1} \frac{3}{5} \Bigg) \cos \Bigg( \cot^{-1} \frac{3}{2} \Bigg) - \sin \Bigg( \sin^{-1} \frac{3}{5} \Bigg) \sin \Bigg( \cot^{-1} \frac{3}{2} \Bigg)$

$\displaystyle = \cos \Bigg( \cos^{-1} \frac{4}{5} \Bigg) \cos \Bigg( \cos^{-1} \frac{3}{\sqrt{13}} \Bigg) - \sin \Bigg( \sin^{-1} \frac{3}{5} \Bigg) \sin \Bigg( \sin^{-1} \frac{2}{\sqrt{13}} \Bigg)$

$\displaystyle = \frac{4}{5} \times \frac{3}{\sqrt{13}} - \frac{3}{5} \times \frac{2}{\sqrt{13}} = \frac{6}{5\sqrt{13}} = \text{ RHS. Hence, proved. }$

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$\displaystyle \text{(iii) LHS } = \tan \Bigg( \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5} \Bigg)$

$\displaystyle = \frac{\tan \Big( \sin^{-1} \frac{5}{13} \Big) + \tan \Big( \cos^{-1} \frac{3}{5} \Big) }{1 - \tan \Big( \sin^{-1} \frac{5}{13} \Big) \tan \Big( \cos^{-1} \frac{3}{5} \Big) }$

$\displaystyle = \frac{\tan \Big( \tan^{-1} \frac{5}{12} \Big) + \tan \Big( \tan^{-1} \frac{4}{3} \Big) }{1 - \tan \Big( \tan^{-1} \frac{5}{12} \Big) \tan \Big( \tan^{-1} \frac{4}{3} \Big) }$

$\displaystyle = \frac{\frac{5}{12}+ \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}} = \frac{63}{16} = \text{ RHS. Hence, proved. }$

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$\displaystyle \text{(iv) LHS } = \sin \Bigg( \cos^{-1} \frac{3}{5} + \sin^{-1} \frac{5}{13} \Bigg)$

$\displaystyle = \sin \Bigg( \cos^{-1} \frac{3}{5} \Bigg) \cos \Bigg( \sin^{-1} \frac{5}{13} \Bigg) + \cos \Bigg( \cos^{-1} \frac{3}{5} \Bigg) \sin \Bigg( \sin^{-1} \frac{5}{13} \Bigg)$

$\displaystyle = \sin \Bigg( \sin^{-1} \frac{4}{5} \Bigg) \cos \Bigg( \cos^{-1} \frac{12}{13} \Bigg) + \cos \Bigg( \cos^{-1} \frac{3}{5} \Bigg) \sin \Bigg( \sin^{-1} \frac{5}{13} \Bigg)$

$\displaystyle = \frac{4}{5} \times \frac{12}{13} + \frac{3}{5} \times \frac{5}{13} = \frac{63}{65} = \text{ RHS. Hence, proved. }$

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$\displaystyle \text{Question 3: Solve: } \cos \Bigg( \sin^{-1} x \Bigg) = \frac{1}{6}$

$\displaystyle \cos \Bigg( \sin^{-1} x \Bigg) = \frac{1}{6}$

$\displaystyle \Rightarrow \cos ( \cos^{-1} \sqrt{1-x^2} ) = \frac{1}{6}$

$\displaystyle \Rightarrow \sqrt{1-x^2} = \frac{1}{6}$

$\displaystyle \Rightarrow 1- x^2 = \frac{1}{36}$

$\displaystyle \Rightarrow 1 - \frac{1}{6} = x^2$

$\displaystyle \Rightarrow x^2 = \frac{35}{36}$

$\displaystyle \Rightarrow x = \pm \frac{\sqrt{35}}{6}$

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$\displaystyle \text{Question 4: Solve: } \cos \Bigg( 2\sin^{-1} (-x) \Bigg) = 0$

$\displaystyle \cos ( 2\sin^{-1} (-x) ) = 0$

$\displaystyle \Rightarrow \cos^2 ( \sin^{-1} (-x)) - \sin^2 ( \sin^{-1}(-x)) = 0$

$\displaystyle \Rightarrow \cos^2 ( \cos^{-1} \sqrt{1-x^2}) - \sin^2 ( \sin^{-1}(-x)) = 0$

$\displaystyle \Rightarrow (\sqrt{1-x^2})^2 - ( -x)^2 = 0$

$\displaystyle \Rightarrow 1 - 2 x^2 = 0$

$\displaystyle \Rightarrow x^2 = \frac{1}{2}$

$\displaystyle \Rightarrow x = \pm \frac{1}{\sqrt{2}}$