Question

If ${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z+{\cos }^{-1}t=4\pi$ , then the value of $x^2+y^2+z^2+t^2$ is

• A. xy + zy + zt
• B. 1 - 2xyzt
• C. 4
• D. 6

Answer 1

Answer: (C)

$co{s}^{-1}x+co{s}^{-1}y+co{s}^{-1}z+co{s}^{-1}z=\pi$
which is only possible when
$co{s}^{-1}x=co{s}^{-1}y+co{s}^{-1}z=co{s}^{-1}t=\pi$
because principal value of $co{s}^{-1}x$ or $co{s}^{-1}y$ or $co{s}^{-1}z$ or $co{s}^{-1}t$ lies in $\left[0,\pi \right]$
$\therefore x=y=z=tcos\pi =-1$
$x^2+y^2+z^2+t^2=4$