Question

If ${\cos }^{-1}p+{\cos }^{-1}q+{\cos }^{-1}r=3\pi$, then $p^2+q^2+r^2+2pqr$ is equal to?

• A. $3$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is B $1$

${\cos }^{-1}p+{\cos }^{-1}q+{\cos }^{-1}r=3\pi$
We know that, if $y={\cos }^{-1}x$, then $-1\le x\le 1$ and $0\le y\le \pi$
Hence, the given equation will hold only when each is $\pi$
$\therefore p=q=r=\cos \pi =-1$
$\therefore p^2+q^2+r^2+2pqr$
$={(-1)}^2+{(-1)}^2+{(-1)}^2+2(-1)(-1)(-1)=3-2=1$