${sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} = \frac{π}{2}$

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Solution

$\begin{matrix} {sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} - \frac{π}{2} l e t {sin}^{- 1} \frac{5}{x} = A {sin}^{- 1} \frac{12}{x} = B sin A \frac{5}{x} sin B = \frac{12}{x} \Rightarrow {sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} = \frac{π}{2} \Rightarrow A + B = \frac{π}{2} \Rightarrow A = \frac{π}{2} - B \Rightarrow sin A = sin (\frac{π}{2} - B) \Rightarrow sin = cos B \Rightarrow sin A = \sqrt{1 - {sin}^{2} B} \Rightarrow \frac{5}{x} = \sqrt{1 - {(\frac{12}{x})}^{2}} \Rightarrow \frac{25}{x^{2}} = 1 - \frac{144}{x^{2}} \Rightarrow \frac{25 + 144}{x^{2}} = 1 \Rightarrow x^{2} = 169 \Rightarrow x = \pm 13 \end{matrix}$

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{sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} = \frac{π}{2}

{sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} = \frac{π}{2}, (x > 0)

x =

s i n^{- 1} \frac{5}{x} + s i n^{- 1} \frac{12}{x} = \frac{π}{2}

x

x

{sin}^{- 1} (\frac{5}{x}) + {sin}^{- 1} (\frac{12}{x}) = \frac{π}{2}

x : {sin}^{- 1} \frac{5}{x} + {sin}^{- 1} \frac{12}{x} = \frac{π}{2}, x \neq 0

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