Question

If ${\cos }^{-1}\left(\frac{2}{3x}\right)+{\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}\left(x>\frac{3}{4}\right)$ then $x$ is equal to :

• A. $\frac{\sqrt{145}}{12}$
• B. $\frac{\sqrt{145}}{10}$
• C. $\frac{\sqrt{146}}{12}$
• D. $\frac{\sqrt{145}}{11}$

Answer 1

Answer: (A)

${\cos }^{-1}\left(\frac{2}{3x}\right)+{\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}\left(x>\frac{3}{4}\right)$
${\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{2}{3x}\right)$
${\cos }^{-1}\left(\frac{3}{4x}\right)={\sin }^{-1}\left(\frac{2}{3x}\right)$
$\cos \left({\cos }^{-1}\left(\frac{3}{4x}\right)\right)=\cos \left({\sin }^{-1}\frac{2}{3x}\right)$
$\frac{3}{4x}=\frac{\sqrt{9x^2-4}}{3x}$
$\frac{81}{16}+4=9x^2$
$x^2=\frac{145}{16\times 9}\Rightarrow x=\frac{\sqrt{145}}{12}$