If cos ^{-1} alpha+cos ^{-1} beta+cos ^{-1} g | vedclass

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(C) Given that $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$. We know that the range of $\cos ^{-1} x$ is $[0, \pi]$. Since the sum of

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$6$

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(C) Given that $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$. We know that the range of $\cos ^{-1} x$ is $[0, \pi]$. Since the sum of three values, each at most $\pi$, is $3 \pi$, each term must be equal to $\pi$. Therefore, $\cos ^{-1} \alpha = \pi$, $\cos ^{-1} \beta = \pi$, and $\cos ^{-1} \gamma = \pi$. This implies $\alpha = \cos(\pi) = -1$, $\beta = \cos(\pi) = -1$, and $\gamma = \cos(\pi) = -1$. Now, we calculate the expression $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$. Substituting $\alpha = -1$, $\beta = -1$, and $\gamma = -1$: $(-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1) = (-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6$.

If $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$, then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ is equal to

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