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If $x>0$ and $\sin^{-1}\bigg(\large{\frac{5}{x}}\bigg)+\sin^{-1}\bigg(\frac{12}{x}\bigg)=\large\frac{\pi}{2},$ than find the value of $x$.

$\begin{array}{1 1} \pm 13 \\ \pm \frac{1}{13} \\ \pm \frac{5}{12} \\ \pm \frac{12}{5} \end{array} $

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  • \( sin^{-1}x=cos^{-1}\sqrt{1-x^2} \)
  • \(cos\theta=sin(\large\frac{\pi}{2}-\theta)\)
given \( sin^{-1}\large\frac{5}{x}+sin^{-1}\large\frac{12}{x}=\large\frac{\pi}{2} \)
\( \Rightarrow sin^{-1}\large\frac{12}{x}=\large\frac{\pi}{2}-sin^{-1}\large\frac{5}{x}\)
take sin on both sides
\( \Rightarrow sin(sin^{-1}\large\frac{12}{x})=sin(\large\frac{\pi}{2}-sin^{-1}\large\frac{5}{x})\)
(By taking \(\theta=sin^{-1}\large\frac{5}{x}\) in the above formula of \(cos\theta\))
\( \Rightarrow \large\frac{12}{x}= cos.sin^{-1}\large\frac{5}{x}=\sqrt{1-\large\frac{25}{x^2}} \)
By taking \(\large\frac{5}{x}\) in the place of x in the above formula of \(sin^{-1}x\),
squaring both the sides
\(\Rightarrow\: \large\frac{144}{x^2}=1-\large\frac{25}{x^2} \)
\(\Rightarrow\: \large\frac{144}{x^2}+\large\frac{25}{x^2}=1 \)
\( \Rightarrow\:\large\frac{169}{x^2}=1 \: or \: x=\pm13\)


answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1

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