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Q.
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
Inverse Trigonometric Functions
Solution:
$cos^{-1}x+cos^{-1}y+cos^{-1}z+cos^{-1}z = \pi $
which is only possible when
$ cos^{-1}x=cos^{-1}y+cos^{-1}z = cos^{-1}t = \pi $
because principal value of $cos^{-1}x$ or $cos^{-1}y$ or $cos^{-1}z$ or $cos^{-1}t$ lies in $\left[0, \pi\right]$
$ \therefore x = y = z = t \, cos\, \pi= -1$
$x^{2}+y^{2}+z^{2}+t^{2}= 4$