Question

If ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}\left(z\right)=3\pi$, then $xy+yz+zx=?$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct option:

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

Given, ${\cos }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}\left(z\right)=3\pi$

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

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

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

Similarly,

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

Step 2. Put the values of $x,y$ and $z$ in given expression:

$\begin{array}{rcl}xy+yz+zx& =& (-1)(-1)+(-1)(-1)+(-1)(-1)\\ & =& 1+1+1\\ & =& \mathbf{}\mathbf{3}\end{array}$

Hence, Option ‘D’ is Correct.