If ${\cos }^{-1}\sqrt{\mathrm{p}}+{\cos }^{-1}\sqrt{1-\mathrm{p}}+{\cos }^{-1}\sqrt{1-\mathrm{q}}=\frac{3\mathrm{\pi }}{4},$ then the value of q is

$$\begin{gathered} {\cos }^{-1}\sqrt{\mathrm{p}}+{\cos }^{-1}\sqrt{1-{\left(\sqrt{\mathrm{p}}\right)}^2}+{\cos }^{-1}\sqrt{1-\mathrm{q}}=\frac{3\mathrm{\pi }}{4} \\ \Rightarrow {\cos }^{-1}\sqrt{\mathrm{p}}+{\sin }^{-1} \sqrt{\mathrm{p}}+{\cos }^{-1}\sqrt{1-\mathrm{q}}=\frac{3\mathrm{\pi }}{4} \\ \Rightarrow \frac{\mathrm{\pi }}{2}+{\cos }^{-1}\sqrt{1-\mathrm{q}}=\frac{3\mathrm{\pi }}{4} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\cos }^{-1}\sqrt{1-\mathrm{q}}=\frac{3\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{2}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \sqrt{1-\mathrm{q}}=\cos \frac{\mathrm{\pi }}{4}\therefore \sqrt{1-\mathrm{q}}=\frac{1}{\sqrt{2}} \\ \Rightarrow 1-\mathrm{q}=\frac{1}{2}\Rightarrow \mathrm{q}=\frac{1}{2} \end{gathered}$$

$\Rightarrow \mathrm{q}=\frac{1}{2}$