Question

If ${\cos }^{-1}\alpha +{\cos }^{-1}\beta$

$+{\cos }^{-1}\gamma =3\pi$, then

$\alpha (\beta +\gamma )+\beta (\gamma +\alpha )+\gamma (\alpha +\beta )$ equals

• A. $6$
• B. $12$
• C. $0$
• D. $1$

Answer 1

The correct option is A $6$

${\cos }^{-1}\alpha +{\cos }^{-1}\beta +{\cos }^{-1}\gamma =3\pi$

$\therefore 0\le {\cos }^{-1}x\le \pi$

$\Rightarrow {\cos }^{-1}\alpha +{\cos }^{-1}\beta +{\cos }^{-1}\gamma =3\pi$

if and only if

${\cos }^{-1}\alpha ={\cos }^{-1}\beta ={\cos }^{-1}\gamma =\pi$

$\Rightarrow \alpha =\beta =\gamma =-1$
$[\because {\cos }^{-1}(-1)=\pi ]$

$\Rightarrow \alpha (\beta +\gamma )+\beta (\gamma +\alpha )+\gamma (\alpha +\beta )$

$=-1(-1-1)-1(-1-1)-1(-1-1)=2+2+2$

$=6$

Therefore, option $\left(c\right)$ is correct.