Question

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

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

Answer 1

The correct option is A $\frac{\sqrt{145}}{12}$

Given : ${\cos }^{-1}\left(\frac{2}{3x}\right)+{\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}\left(\frac{2}{3x}\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{3}{4x}\right)$
$\Rightarrow {\cos }^{-1}\left(\frac{2}{3x}\right)={\sin }^{-1}\left(\frac{3}{4x}\right)[\because {\sin }^{-1}y+{\cos }^{-1}y=\frac{\pi }{2}]$
$\Rightarrow {\cos }^{-1}\left(\frac{2}{3x}\right)={\cos }^{-1}\left(\sqrt{1-{\left(\frac{3}{4x}\right)}^2}\right)$
Applying $\cos$ both sides, we get
$\frac{2}{3x}=\sqrt{1-{\left(\frac{3}{4x}\right)}^2}$
$\Rightarrow \frac{4}{9x^2}=1-{\left(\frac{3}{4x}\right)}^2$
$\Rightarrow \frac{4}{9x^2}+\frac{9}{16x^2}=1$
$\Rightarrow \frac{64+81}{9\times 16\times x^2}=1$
$\Rightarrow x=\frac{\sqrt{145}}{12}[\because x>\frac{3}{4}]$