Question

if cos-1x + cos-1y + cos-1z = pi


prove that x2 + y2 + z2 +2xyz = 1

Answer 1

I hope this answer helps you

Answer 2

$\bold{x^{2}+y^{2}+z^{2}+2 x y z=1}$ is proven.

Solution:

Let us first transfer $\cos ^{-1} z$ left side of the equal to π to convert it into $\pi-\cos ^{-1} z$ which will result in $\cos ^{-1}(-z)$.

$\begin{array}{l}{\cos ^{-1} x+\cos ^{-1} y=\pi-\cos ^{-1} z} \\ \\{\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}(-z)}\end{array}$

Now replace  

$A=\cos ^{-1} x ; B=\cos ^{-1} y$ y so as bring the equation in terms of x, y, z so to make it equal with the RHS

$\begin{array}{l}{\cos A=x ; \cos B=y} \\ \\{\sin A=\sqrt{1-x^{2}} ; \sin B=\sqrt{1-y^{2}}} \\ \\{\cos (A+B)=\cos A \cos B-\sin A \sin B} \\ \\{\cos (A+B)=x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}} \\ \\{A+B=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)}\end{array}$

Placing $A+B=\cos ^{-1}(-z) because \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}(-z)$, as cosA=x; cosB=y  

$\begin{array}{l}{A+B=\cos ^{-1}(-z)} \\ \\{\cos ^{-1}(-z)=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)} \\ \\{-z=x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}}\end{array}$

$\begin{array}{l}{z+x y=+\sqrt{1-x^{2}} \sqrt{1-y^{2}}} \\ \\{(z+x y)^{2}=\left(1-x^{2}\right)\left(1-y^{2}\right)}\end{array}$

After solving the equation we simplify it in saturated form we get the RHS, therefore  

$x^{2}+y^{2}+z^{2}+2 x y z=1$

Hence Proved.