If cos^{-1} sqrt{p} + cos^{-1} sqrt{1-p} + co | vedclass

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(D) Let $\alpha = \cos^{-1} \sqrt{p},$ $\beta = \cos^{-1} \sqrt{1-p},$ and $\gamma = \cos^{-1} \sqrt{1-q}.$Then $\cos \alpha = \sqrt{p},$ $\cos \beta

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$\frac{1}{2}$

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(D) Let $\alpha = \cos^{-1} \sqrt{p},$ $\beta = \cos^{-1} \sqrt{1-p},$ and $\gamma = \cos^{-1} \sqrt{1-q}.$Then $\cos \alpha = \sqrt{p},$ $\cos \beta = \sqrt{1-p},$ and $\cos \gamma = \sqrt{1-q}.$Consequently,$\sin \alpha = \sqrt{1-p},$ $\sin \beta = \sqrt{p},$ and $\sin \gamma = \sqrt{q}.$The given equation is $\alpha + \beta + \gamma = \frac{3\pi}{4}.$This implies $\alpha + \beta = \frac{3\pi}{4} - \gamma.$Taking cosine on both sides: $\cos(\alpha + \beta) = \cos\left(\frac{3\pi}{4} - \gamma\right).$$\cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos\left(\pi - \left(\frac{\pi}{4} + \gamma\right)\right) = -\cos\left(\frac{\pi}{4} + \gamma\right).$$\sqrt{p} \sqrt{1-p} - \sqrt{1-p} \sqrt{p} = -\left(\frac{1}{\sqrt{2}} \cos \gamma - \frac{1}{\sqrt{2}} \sin \gamma\right).$$0 = -\frac{1}{\sqrt{2}}(\sqrt{1-q} - \sqrt{q}).$Thus,$\sqrt{1-q} = \sqrt{q}.$Squaring both sides,$1-q = q,$ which gives $2q = 1,$ or $q = \frac{1}{2}.$

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

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