Question

If $co{s}^{-1}x+co{s}^{-1}y+co{s}^{-1}z=\pi$, then

• A. $x^2+y^2+z^2+xyz=0$
• B. $x^2+y^2+z^2+2xyz=0$
• C. $x^2+y^2+z^2+xyz=1$
• D. $x^2+y^2+z^2+2xyz=1$

Answer 1

Answer: (D)

$x^2+y^2+z^2+2xyz=1$

$co{s}^{-1}\left(x\right)+co{s}^{-1}\left(y\right)=\pi -co{s}^{-1}\left(z\right)$
$co{s}^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})=\pi -co{s}^{-1}\left(z\right)$
Taking cos on both the sides, we get
$xy-\sqrt{1-x^2}\sqrt{1-y^2}=-z$
$xy+z=\sqrt{1-x^2}\sqrt{1-y^2}$
Squaring on both the sides, we get
$x^2{y}^2+z^2+2xyz=1-y^2-x^2+x^2{y}^2$
$x^2+y^2+z^2+2xyz=1$