If ${\cos }^{-1}p+{\cos }^{-1}q+{\cos }^{-1}r=\pi$ then prove that $p^2+q^2+r^2+2pqr=1$

Given ${\cos }^{-1}p+{\cos }^{-1}q+{\cos }^{-1}r=\pi$

${\cos }^{-1}p+{\cos }^{-1}q=\pi -{\cos }^{-1}r$

${\cos }^1\left(pq-\sqrt{1-p^2}\sqrt{1-q^2}\right)=\pi -{\cos }^{1}r$

${\cos }^1\left(pq-\sqrt{1-p^2}\sqrt{1-q^2}\right)={\cos }^1(-r)$

$pq-\sqrt{1-p^2}\sqrt{1-q^2}=-r$

$pq+r=\sqrt{1-p^2}\sqrt{1-q^2}$

S.O.B.S.

$(pq+r{)}^2={\left(\sqrt{1-p^2}\sqrt{1-q^2}\right)}^2$

$p^2{q}^2+2pqr+r^2=\left(1-p^2\right)\left(1-q^2\right)$

$p^2{q}^2+2pqr+r^2=1-p^2-q^2+p^2{q}^2$

$\because$$p^2+q^2+r^2+2pqr=1$