Q.(C) cos−1(3𝑥) (D) 3 cos−1 𝑥 1 h

Domain of Inverse Secant

To define $\sec^{-1} x$ we ask: for which values of $x$ does the equation $\sec\theta = x$ have a solution? The answer is the domain of inverse secant, and it looks quite different from the domain of $\sin^{-1}$ or $\cos^{-1}$.

Why $|x| \ge 1$

Recall $\sec\theta = \dfrac{1}{\cos\theta}$, and $\cos\theta$ always lies in $[-1, 1]$. Taking reciprocals:

• when $|\cos\theta| \le 1$, we get $|\sec\theta| \ge 1$.

So secant never outputs a value strictly between $-1$ and $1$. There is simply no angle whose secant is, say, $0.5$. Therefore

The interval $(-1, 1)$ is excluded — this is the single most-tested fact about inverse secant.

The matching range

Like every trig function, secant repeats, so we must restrict it to make it one-to-one before inverting. The conventional principal-value choice keeps $\theta$ in

We remove $\theta = \tfrac{\pi}{2}$ because $\cos\tfrac{\pi}{2} = 0$, so $\sec\tfrac{\pi}{2}$ is undefined. On $[0, \tfrac{\pi}{2})$ secant runs from $1$ up to $+\infty$, covering $[1, \infty)$; on $(\tfrac{\pi}{2}, \pi]$ it runs from $-\infty$ up to $-1$, covering $(-\infty, -1]$. Together these give exactly $|x|\ge 1$ — matching the domain above. …