Question

If $co{s}^{-1}\alpha +co{s}^{-1}\beta +co{s}^{-1}\gamma =3\pi$, 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)

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