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

We have, $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Using the identity $\cos^{-1} \sqrt{1 - p} = \sin^{-1} \sqrt{p}$:$\cos^{-1} \sqrt{p} + \sin^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Since $\cos^{-1} \theta + \sin^{-1} \theta = \frac{\pi}{2}$:$\frac{\pi}{2} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$$\cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4} - \frac{\pi}{2}$$\cos^{-1} \sqrt{1 - q} = \frac{\pi}{4}$$\sqrt{1 - q} = \cos\left(\frac{\pi}{4}\right)$$\sqrt{1 - q} = \frac{1}{\sqrt{2}}$Squaring both sides:$1 - q = \frac{1}{2}$$q = 1 - \frac{1}{2}$$q = \frac{1}{2}$