If \(\cos ^{-1} \sqrt{\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{q}}=\frac{3 \pi}{4}\), then \(\mathrm{q}\)

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