Question

cos inverse x + cos inverse y + cos inverse z =π ,then prove that x square + y square + z square + 2xyz= 1

Answer 1

Check the attachment
after simplifying it you will get the answer

Answer 2

$\bold{x^{2}+y^{2}+z^{2}+2 x y z=1 \text { if } \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi}$

Given :  

$\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi$

To find :    

If  $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi then  x^{2}+y^{2}+z^{2}+2 x y z=1$

Solution:  

$\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi$

Taking $\cos ^{-1} z$ to other side  

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

(We know that $\cos ^{-1} a+\cos ^{-1} b=\cos ^{-1}\left(a b-o r+\sqrt{1-a^{2}} \sqrt{1-b^{2}}\right)$

Apply cos trigonometric function to both sides  

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

\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)=-\cos \left(\cos ^{-1} z\right)

(As cos is negative in second quadrant, $\cos (\pi-x)=-\cos x$ )

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

Squaring on both sides  

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

$\bold{x^{2}+y^{2}+z^{2}+2 x y z=1 \text { if } \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi}$

Hence proved.