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

$\begin{array}{c}\text{ Let }{\cos }^{-1}\sqrt{p}+{\cos }^{-1}\sqrt{1-p}+{\cos }^{-1}\sqrt{1-q}=\frac{3\pi }{4}\\ \text{ Let }a={\cos }^{-1}\sqrt{p},b={\cos }^{-1}\sqrt{1-p}\text{ and }c={\cos }^{-1}\sqrt{1-q}\\ \Rightarrow \cos a=\sqrt{p},\cos b=\sqrt{1-p},\cos c=\sqrt{1-q}\\ \Rightarrow {\cos }^{2}a=p,{\cos }^{2}b=1-p,{\cos }^{2}c=1-q\\ \text{ Now, }{\sin }^{2}a=1-{\cos }^{2}a=1-p\\ \Rightarrow \sin a=\sqrt{1-p}\\ {\sin }^{2}b=1-{\cos }^{2}b=1-1+p\Rightarrow \sin b=\sqrt{p}\\ {\sin }^{2}c=1-{\cos }^{2}c=1-1+q\Rightarrow \sin c=\sqrt{q}\\ We have a+b+c=\frac{3\pi }{4}\Rightarrow a+b=\frac{3\pi }{4}-c\\ \sin \left(a+b\right)=\sin \left(\frac{3\mathrm{\pi }}{4}-c\right)\\ \sin a\cos b+\cos a\sin b=\sin \left(\frac{3\mathrm{\pi }}{4}-c\right)\\ \Rightarrow {\sin }^{2}a+{\cos }^{2}a=\sin \left(\frac{3\mathrm{\pi }}{4}-c\right) \left(\because \sin a=\cos b, \cos a=\sin b\right)\\ \Rightarrow 1=\sin \left(\frac{3\mathrm{\pi }}{4}-c\right)\\ \Rightarrow \frac{3\mathrm{\pi }}{4}-c=\frac{\mathrm{\pi }}{2}\\ \Rightarrow c=\frac{\mathrm{\pi }}{4}\\ \therefore q={\sin }^{2}c=\frac{1}{2}=0.5\end{array}$