Question

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

• A. $\frac{\sqrt{145}}{11}$
• B. $\frac{\sqrt{145}}{12}$
• C. $\frac{\sqrt{145}}{13}$
• D. $\frac{\sqrt{145}}{14}$

Answer 1

The correct option is B

$\frac{\sqrt{145}}{12}$

Find the value of $x$:

Given, ${\cos }^{-1}\left(\frac{2}{3x}\right)+{\cos }^{-1}\left(\frac{3}{4x}\right)=\frac{\pi }{2}$

$\Rightarrow$ ${\cos }^{-1}\left(\frac{2}{3x}\right)=\frac{\pi }{2}-co{s}^{-1}\left(\frac{3}{4x}\right)$

$\Rightarrow$ ${\cos }^{-1}\left(\frac{2}{3x}\right)=co{s}^{-1}\left(\frac{\sqrt{16x^2-9}}{4x}\right)$

$\Rightarrow$ $\left(\frac{2}{3x}\right)=\left(\frac{\sqrt{16x^2-9}}{4x}\right)$

$\Rightarrow$ $\sqrt{16x^2-9}=\frac{8}{3}$

$\Rightarrow$ $16x^2-9=\frac{64}{9}$

$\Rightarrow$ $x^2=\frac{145}{(9\times 16)}$

$\mathbf{\therefore }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\sqrt{\mathbf{145}}}{\mathbf{12}}$

Hence, Option ‘B’ is Correct.