Inverse Trigonometric Function

$\dfrac {\arcsin(1-\sqrt { x } ) }{\arccos(1-\sqrt { x } ) } =1$

$\color{#3D99F6}\text{Now Cross Multiplying}$

$\implies \sin ^{-1} (1-\sqrt{x}) = \cos^{-1} (1- \sqrt{x})$

$\implies \sin ^{-1} (1-\sqrt{x}) = \dfrac{\pi}{2} - \sin ^{-1} (1-\sqrt{x})$

$\implies 2\sin ^{-1} (1-\sqrt{x}) = \dfrac{\pi}{2}$

$\implies \sin ^{-1} (1-\sqrt{x}) = \dfrac{\pi}{4}$

$\implies 1- \sqrt{x} = \dfrac{1}{\sqrt{2}}$

$\implies 1 - \dfrac{1}{\sqrt{2}} = \sqrt{x}$

$\implies \dfrac{\sqrt{2}-1}{\sqrt{2}} = \sqrt{x}$

$\implies x = \dfrac{3-2\sqrt{2}}{2}$

Now Putting the value of $x$ in the required expression,

$\implies \tan^{-1} \sqrt{x^2 - 3x + \dfrac{13}{4}} \implies \tan^{-1} \sqrt{\dfrac{9+8-12\sqrt{2}}{4} - \bigg(\dfrac{9-6\sqrt{2}}{2} \bigg) + \dfrac{13}{4}}$

$\implies \tan^{-1} \sqrt{\dfrac{17-12\sqrt{2}-18+12\sqrt{2}}{4} + \dfrac{13}{4}}$ $\implies \tan^{-1} \sqrt{\dfrac{12}{4}} \implies \tan^{-1}{\sqrt{3}} \implies \boxed{\dfrac{\pi}{3}}$ [ $\color{#D61F06} \text{OUR PRETTY ANSWER}$ ]