[-1, 1] is the domain for which of the following inverse trigonometric functions?

The correct option is sin^-1x.

To understand why, let us analyze the domain of the inverse trigonometric functions mentioned in the options.
Recall that if a trigonometric function y = f(x) is defined on a domain and we restrict it to make it one-to-one and onto (bijective), we can define its inverse function x = f^-1(y). The domain of the inverse function f^-1(y) is the range of the restricted trigonometric function f(x).

Let us look at each option individually:

1. sin^-1x (Inverse Sine Function):
The original sine function, y = sin(θ), outputs values between -1 and 1, inclusive, for all real numbers θ. That is, its range is [-1, 1].
When we restrict the domain of sin(θ) to $[-\frac{\pi }{2},\frac{\pi }{2}]$ to make it invertible, the range remains $[-1,1]$.
Therefore, the domain of its inverse, y = sin^-1x, is:
$\text{Domain}=[-1,1]$
This matches the interval given in the question.

2. cot^-1x (Inverse Cotangent Function):
The cotangent function, y = cot(θ), takes any value in the set of real numbers. Its range is $(-\infty ,\infty )$ (or $ℝ$).
Consequently, the domain of y = cot^-1x is:
$\text{Domain}=(-\infty ,\infty )$

3. tan^-1x (Inverse Tangent Function):
Similar to the cotangent function, the tangent function, y = tan(θ), has a range of all real numbers, $(-\infty ,\infty )$.
Thus, the domain of y = tan^-1x is:
$\text{Domain}=(-\infty ,\infty )$

4. sec^-1x (Inverse Secant Function):
The secant function, y = sec(θ), takes values that are either greater than or equal to 1, or less than or equal to -1. Its range is $(-\infty ,-1]\cup [1,\infty )$.
Thus, the domain of y = sec^-1x is:
$\text{Domain}=(-\infty ,-1]\cup [1,\infty )$

Comparing all the domains, we see that $[-1,1]$ is strictly the domain of sin^-1x (and cos^-1x, which is not listed in the options).