Question

If ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}\left(z\right)+{\cos }^{-1}\left(t\right)=4\pi$, 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

The correct option is C

$4$

Explanation for the correct option:

Step 1. Find the value of ${\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{y}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{z}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{t}}^{\mathbf{2}}$:

Given, ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}\left(z\right)+{\cos }^{-1}\left(t\right)=4\pi$

As we know, $0\le {\cos }^{-1}\left(x\right)\le \pi$

Thus, maximum value of ${\cos }^{-1}\left(x\right)=\pi$ satisfies the given equation

$\begin{array}{rcl}x& =& \cos \pi \\ & =& -1\end{array}$

Similarly,

$\begin{array}{rcl}y& =& z\\ & =& t\\ & =& -1\end{array}$

Step 2.Put the value of $x,y,z$ and $t$ in given expression, we get

$\begin{array}{rcl}{x}^2+y^2+z^2+t^2& =& {(-1)}^2+{(-1)}^2+{(-1)}^2+{(-1)}^2\\ & =& 1+1+1+1\\ & =& \mathbf{4}\end{array}$

Hence, Option ‘C’ is Correct.