Question

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

Answer 1

Answer: 133

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