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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 $x$.

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

Toolbox:
• $sin^{-1}x=cos^{-1}\sqrt{1-x^2}$
• $cos\theta=sin(\large\frac{\pi}{2}-\theta)$
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 $sin^{-1}x$,
$sin^{-1}\large\frac{5}{x}=cos^{-1}\sqrt{1-\large\frac{25}{x^2}}$
$\large\frac{12}{x}=cos(sin^{-1}\large\frac{5}{x})=cos\big(cos^{-1}\sqrt{1-\large\frac{25}{x^2}}\big)$
$\Rightarrow\:\large\frac{12}{x}=\sqrt{1-\large\frac{25}{x^2}}$

squaring both the sides
$\Rightarrow\: \large\frac{144}{x^2}=1-\large\frac{25}{x^2}$
$\Rightarrow\: \large\frac{144}{x^2}+\large\frac{25}{x^2}=1$
$\Rightarrow\:\large\frac{169}{x^2}=1 \: or \: x=\pm13$

edited Mar 19, 2013