Question

Range of sec^2x

Answer 1

[1, \infty)

Answer 2

We use the trigonometric identity: $\sec^2(x) = 1 + \tan^2(x)$ The range of the tangent function, $\tan(x)$, is $(-\infty, \infty)$. When we square $\tan(x)$, the range of $\tan^2(x)$ becomes $[0, \infty)$, as the square of any real number is non-negative. Therefore, for $\sec^2(x) = 1 + \tan^2(x)$, the minimum value occurs when $\tan^2(x) = 0$, which gives $\sec^2(x) = 1 + 0 = 1$. As $\tan^2(x)$ can take any non-negative value up to infinity, $\sec^2(x)$ can take any value from $1$ up to infinity. Thus, the range of $\sec^2(x)$ is $[1, \infty)$.