# 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