$\begin{aligned} & \cos ^{-1} \sqrt{\mathrm{p}}-\cos ^{-1} \sqrt{1-\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{q}}=\frac{3 \pi}{4} \\ & \text { Let } \mathrm{t}=\cos ^{-1} \sqrt{\mathrm{p}} \\ & \Rightarrow \mathrm{p}=\cos ^2 \mathrm{t} \\ & \Rightarrow \mathrm{p}=1-\sin ^2 \mathrm{t} \\ & \Rightarrow \sin \mathrm{t}=\sqrt{1-\mathrm{p}} \\ & \Rightarrow \mathrm{t}=\sin ^{-1} \sqrt{1-\mathrm{p}}\end{aligned}$
$\Rightarrow \cos ^{-1} \sqrt{\mathrm{p}}=\sin ^{-1} \sqrt{1-\mathrm{p}}$
$\therefore \quad$ Given equation becomes
$\sin ^{-1} \sqrt{1-p}-\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$
$\therefore \quad \frac{\pi}{2}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4} \quad \ldots\left[\because \cos ^{-1} a+\sin ^{-1} a=\frac{\pi}{2}\right]$
$\begin{array}{ll}\therefore & \cos ^{-1} \sqrt{1-q}=\frac{-\pi}{4} \\ \therefore & \sqrt{1-q}=\cos \left(-\frac{\pi}{4}\right) \\ \therefore & q=1-\frac{1}{2} \\ \therefore & q=\frac{1}{2}\end{array}$