If cos^{-1} p + cos^{-1} q + cos^{-1} r = pi, | vedclass

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(B) Given that $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$. Let $\cos^{-1} p = A$,$\cos^{-1} q = B$,and $\cos^{-1} r = C$. Then $A + B + C = \pi$,

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(B) Given that $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$. Let $\cos^{-1} p = A$,$\cos^{-1} q = B$,and $\cos^{-1} r = C$. Then $A + B + C = \pi$,which implies $A + B = \pi - C$. Taking $\cos$ on both sides: $\cos(A + B) = \cos(\pi - C) = -\cos C$. Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get: $pq - \sqrt{1 - p^2} \sqrt{1 - q^2} = -r$. $pq + r = \sqrt{1 - p^2} \sqrt{1 - q^2}$. Squaring both sides: $(pq + r)^2 = (1 - p^2)(1 - q^2)$. $p^2q^2 + r^2 + 2pqr = 1 - p^2 - q^2 + p^2q^2$. $p^2 + q^2 + r^2 + 2pqr = 1$.

If $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$,then $p^2 + q^2 + r^2 + 2pqr = $

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