What is the value of cos-1(-x) for all x belongs to [-1, 1]?

The correct option is π �� cos⁻¹(x).

To understand why this relationship holds, let us derive it step-by-step using the definition of inverse trigonometric functions.
Let:
$$y={\cos }^{-1}(-x)$$

By definition of the inverse cosine function, for $x\in [-1,1]$, the value of $y$ must lie in the principal value branch, which is:
$$y\in [0,\pi ]$$

From the equation $y={\cos }^{-1}(-x)$, we can rewrite this by taking the cosine of both sides:
$$\cos \left(y\right)=-x$$

Multiplying both sides by -1 gives:
$$x=-\cos \left(y\right)$$

Using the trigonometric identity $-\cos \left(\theta \right)=\cos (\pi -\theta )$ which represents the second quadrant where cosine is negative, we can substitute it into our equation:
$$x=\cos (\pi -y)$$

Now, taking the inverse cosine of both sides:
$${\cos }^{-1}\left(x\right)=\pi -y$$

Rearranging the terms to solve for $y$:
$$y=\pi -{\cos }^{-1}\left(x\right)$$

Substituting our original assumption for $y$ back into the equation yields the final formula:
$${\cos }^{-1}(-x)=\pi -{\cos }^{-1}\left(x\right)$$