Question

If $cos^{-1}\alpha + cos^{-1}\beta + cos^{-1}\gamma = 3\pi$, then $ \alpha\left(\beta+\gamma\right) + \beta\left(\gamma+\alpha\right)+\gamma\left(\alpha + \beta\right)$ equals

• A. $0$
• B. $1$
• C. $6$
• D. $12$

Answer 1

Answer: (C)

$cos^{-1}\alpha + cos^{-1}\beta + cos^{-1}\gamma = 3\pi$
$\because 0 \le cos^{-1}x \le \pi$
$\Rightarrow cos^{-1}\alpha = cos^{-1}\beta = cos^{-1}\gamma = \pi$
$\Rightarrow \alpha = \beta = \gamma = -1$
$\because \alpha\left(\beta+\gamma\right) + \beta\left(\gamma+\alpha\right)+\gamma\left(\alpha + \beta\right)$
$=-1 \left( -1 -1\right) + \left(-1\right)\left(_{ }-1 - 1\right) + \left(- 1\right)\left(- 1 - 1\right)$
$= 2 + 2 + 2=6$