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

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(A) Given equation: $\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ Let $t=\cos ^{-1} \sqrt{p}$. Then $\sqrt{p}=\cos

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$\frac{1}{2}$

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(A) Given equation: $\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ Let $t=\cos ^{-1} \sqrt{p}$. Then $\sqrt{p}=\cos t$,so $p=\cos ^2 t$. Thus,$1-p=1-\cos ^2 t=\sin ^2 t$,which implies $\sqrt{1-p}=\sin t$. Therefore,$t=\sin ^{-1} \sqrt{1-p}$. Since $t=\cos ^{-1} \sqrt{p}$,we have $\cos ^{-1} \sqrt{p}=\sin ^{-1} \sqrt{1-p}$. Substituting this into the given equation: $\sin ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ Using the identity $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$,we get: $\frac{\pi}{2}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ $\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}-\frac{\pi}{2}=\frac{\pi}{4}$ $\sqrt{1-q}=\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$ Squaring both sides: $1-q=\frac{1}{2}$ $q=1-\frac{1}{2}=\frac{1}{2}$

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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