Question

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

Answer 1

$cos^{-1}\left(\frac{2}{3x}\right)+cos^{-1} \left(\frac{3}{4x}\right)=\frac{\pi}{2} ; \left(x>\,\frac{3}{4}\right)$
$\Rightarrow cos^{-1}\left(\frac{2}{3x}\right)=\frac{\pi}{2}-cos^{-1} \left(\frac{3}{4x}\right)$
$\Rightarrow cos^{-1}\left(\frac{2}{3x}\right)=sin^{-1} \left(\frac{3}{4x}\right)$
$\left[\because sin^{-1}\,x +cos^{-1} x= \frac{\pi}{2}\right]$
put $\Rightarrow sin^{-1} \left(\frac{3}{4x}\right)=\theta $
$\Rightarrow sin\,\theta = \frac{3}{4x}$
$\Rightarrow cos \,\theta=\sqrt{1-sin^{2}\,\theta}=\sqrt{1-\frac{9}{16x^{2}}}$
$\Rightarrow \theta=cos^{-1} \left(\frac{\sqrt{16x^{2}-9}}{4x}\right)$
$\therefore cos^{-1}\left(\frac{2}{3x}\right)=cos^{-1} \left(\frac{\sqrt{16x^{2}-9}}{4x}\right)$
$\Rightarrow \frac{2}{3x}=\frac{\sqrt{16x^{2}-9}}{4x}$
$\Rightarrow x^{2}=\frac{64+81}{9\times16} $
$\Rightarrow x= \pm\sqrt{\frac{145}{144}}$
$\Rightarrow x= \frac{\sqrt{145}}{12}$
$ \Rightarrow \frac{\sqrt{p}}{q}=\frac{\sqrt{145}}{12} \left(\because x >\, \frac{3}{4}\right)$
$\Rightarrow p=145, q=12$
Then, $p-q=145-12=133$