Prove that: $\sin^{- 1} (\frac{4}{5}) + \sin^{- 1} (\frac{5}{13}) + \sin^{- 1} (\frac{16}{65}) = \frac{π}{2}$

Solution

$To prove : \sin^{- 1} (\frac{4}{5}) + \sin^{- 1} (\frac{5}{13}) + \sin^{- 1} (\frac{16}{65}) = \frac{π}{2} Let \sin^{- 1} (\frac{4}{5}) = x \Rightarrow sinx = \frac{4}{5} \Rightarrow cosx = \sqrt{1 - \sin^{2} x} = \frac{3}{5} \sin^{- 1} (\frac{5}{13}) = y \Rightarrow siny = \frac{5}{13} \Rightarrow cosy = \sqrt{1 - \sin^{2} y} = \frac{12}{13} \sin^{- 1} (\frac{16}{65}) = z \Rightarrow sinz = \frac{16}{65} \Rightarrow cosz = \sqrt{1 - \sin^{2} z} = \frac{63}{65} tanx = \frac{4}{3}, tany = \frac{5}{12}, tanz = \frac{16}{63} tanz = \frac{16}{63} \Rightarrow cotz = \frac{63}{16} . . . . . . . . . (i)$

$\tan (x + y) = \frac{\tan (x + y)}{1 - tanx . tany} \Rightarrow \tan (x + y) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{20}{36}} \Rightarrow \tan (x + y) = \frac{63}{16} \Rightarrow \tan (x + y) = cotz . . . . . . . [from equation (i)] \Rightarrow \tan (x + y) = \tan (\frac{π}{2} - z) \Rightarrow x + y = \frac{π}{2} - z \Rightarrow x + y + z = \frac{π}{2} ∴ \sin^{- 1} (\frac{4}{5}) + \sin^{- 1} (\frac{5}{13}) + \sin^{- 1} (\frac{16}{65}) = \frac{π}{2}$

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Inverse Trigonometric Functions

Prove that: $\sin^{- 1} (\frac{4}{5}) + \sin^{- 1} (\frac{5}{13}) + \sin^{- 1} (\frac{16}{65}) = \frac{π}{2}$

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Prove that: $\sin^{- 1} (\frac{4}{5}) + \sin^{- 1} (\frac{5}{13}) + \sin^{- 1} (\frac{16}{65}) = \frac{π}{2}$