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

$\frac{1}{2}$

Clearly, left hand side of the given equation is meaningful if $0 ≤ p ≤ 1$ and $0 ≤ q ≤ 1$.

Let $ p = cos^2\theta .$ Then,

$ cos^{-1}\sqrt{p} + cos^{-1} \sqrt{1-p} +cos^{-1} \sqrt{1-q} =\frac{3\pi}{4}$

$⇒ cos^{-1}(cos \theta ) +  cos^{-1} (sin \theta ) +  cos^{-1} \sqrt{1-q} =\frac{3\pi}{4}$

$⇒ \theta + \frac{\pi}{2} - \theta + cos^{-1} \sqrt{1-q} =\frac{3\pi}{4}$

$⇒cos^{-1} \sqrt{1-q} =\frac{\pi}{4}$

$⇒  \sqrt{1-q} =\frac{1}{\sqrt{2}} ⇒ 1 - q = \frac{1}{2} ⇒ q = \frac{1}{2}$