If ^ - 1 ((2)/(3x)) + ^ - 1 ((3)/(4x)) = (π)/(2)(x > (3)/(4)), then...

$∴ \cos^{- 1}\dfrac{3}{4x} = \sin^{- 1}\dfrac{\sqrt{16{x}^{2} - 9}}{4x}$

Put in equation \left(i\right)$we get, <br /><br />$⇒ \cos^{- 1}\dfrac{2}{3x} + \sin^{- 1}\dfrac{\sqrt{16{x}^{2} - 9}}{4x} = \dfrac{\pi}{2} . . . $\left(ii\right)$

We know $\sin^{- 1}x + \cos^{- 1}x = \dfrac{\pi}{2}$

From equation \left(ii\right)$we get, <br /><br />$⇒ \dfrac{2}{3x} = \dfrac{\sqrt{16{x}^{2} - 9}}{4x}$<br /><br />$⇒ {\left(\dfrac{8}{3}\right)}^{2} = 16{x}^{2} - 9\left(∵ x \ne 0\right)$<br /><br />$⇒ 16{x}^{2} = \dfrac{64}{9} + 9 = \dfrac{145}{9}$<br /><br />$⇒ {x}^{2} = \dfrac{145}{144}$<br /><br />$⇒ x = \dfrac{\sqrt{145}}{12}