Prove the following

Question:

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

then $x$ is equal to:

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

  2. $\frac{\sqrt{145}}{10}$

  3. $\frac{\sqrt{146}}{12}$

  4. $\frac{\sqrt{145}}{11}$


Correct Option: 1

Solution:

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

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

$\cos ^{-1}\left(\frac{3}{4 x}\right)=\sin ^{-1}\left(\frac{2}{3 x}\right)$

$\cos \left(\cos ^{-1}\left(\frac{3}{4 x}\right)\right)=\cos \left(\sin ^{-1} \frac{2}{3 x}\right)$

$\frac{3}{4 x}=\frac{\sqrt{9 x^{2}-4}}{3 x}$

$\frac{81}{16}+4=9 x^{2}$

$x^{2}=\frac{145}{16 \times 9} \Rightarrow x=\frac{\sqrt{145}}{12}$

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