Q.If cos^-1 x + cos^-1 y + cos^-1 z = π, then show that x² + y² + z² + 2xyz = 1.

Set $A=\cos^{-1}x$, $B=\cos^{-1}y$; use $A+B+C=\pi$ to write $z=\cos C$ as $-\cos(A+B)$, then eliminate $A,B$ using $\sin^2A=1-x^2$ etc.

Let $A=\cos^{-1}x$, $B=\cos^{-1}y$, $C=\cos^{-1}z$, so $x=\cos A$, $y=\cos B$, $z=\cos C$, and $A+B+C=\pi$.

From $A+B+C=\pi$, we get $C = \pi-(A+B)$, so:

$z = \cos C = \cos(\pi-(A+B)) = -\cos(A+B) = -(\cos A\cos B - \sin A\sin B) = \sin A\sin B - xy$

Rearranging: $\sin A \sin B = z+xy$.

Squaring both sides:

$\sin^2A\sin^2B = (z+xy)^2$

Using $\sin^2A = 1-\cos^2A = 1-x^2$ and similarly $\sin^2B=1-y^2$: …