FRAC{3\pi }{4}\] .. (i) Let \[a={{\cos }^{-1}}\sqrt{p}b={{\cos }^{-1}}\sqrt{1-p}\] and \[c={{\cos }^{-1}}\sqrt{1-q}\] \[\Rightarrow \cos a=\sqrt{p},\cos b=\sqrt{1-p},\cos c=\sqrt{1-q}\] \[\Rightarrow {{\cos }^{2}}a=p,{{\cos }^{2}}b=1-p,{{\cos }^{2}}c=1-q\] Now, \[{{\sin }^{2}}a=1-{{\cos }^{2}}a=1-p\] \[\Rightarrow \sin a=\sqrt{1-p},\] \[{{\sin }^{2}}b=1-{{\cos }^{2}}b=1-1+p\Rightarrow \sin b=\sqrt{p}\] \[{{\sin }^{2}}c=1-{{\cos }^{2}}c=1-1+q=q\Rightarrow \sin c=\sqrt{q}\] \[\THEREFORE \] equation (i) can be written as \[a+b+c=\frac{3\pi }{4}\Rightarrow a+b=\frac{3\pi }{4}-c\] Take cos on each side, we get \[\cos (a+b)=cos\left( \frac{3\pi }{4}-c \RIGHT)\] \[\Rightarrow \cos a\,\cos b-\sin a\,\sin b\] \[=\cos \left\{ \pi -\left( \frac{\pi }{4}+c \right) \right\}=-\cos \left( \frac{\pi }{4}+c \right)\] Put VALUES of \[\cos a,cosb\] and \[\sin a,\sin b,\] we get \[\sqrt{p}.\sqrt{1-p}-\sqrt{1-p}\sqrt{p}\] \[=-\left( \frac{1}{\sqrt{2}}\sqrt{1-q}-\frac{1}{\sqrt{2}}\sqrt{q} \right)\] \[\Rightarrow 0=\sqrt{1-q}-\sqrt{q}\Rightarrow \sqrt{1-q}=\sqrt{q}\] Squaring on both side;\[\Rightarrow 1-q=q\] \[\Rightarrow 1=2q\Rightarrow q=\frac{1}{2}\]