If , , then the smallest interval in which lies is:

(d) $\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$ Given $x \ge 0$, since $\sin^{-1}x$ and $\cos^{-1}x$ require $x \in [-1, 1]$, the valid domain is $0 \le x \le 1$. For all $x \in [0, 1]$, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$. Thus, $\theta = \frac{\pi}{2} - \tan^{-1}x$. Since $0 \le x \le 1$, we have $0 \le \tan^{-1}x \le \frac{\pi}{4}$. Multiplying by $-1$ gives $-\frac{\pi}{4} \le -\tan^{-1}x \le 0$. Adding $\frac{\pi}{2}$ gives $\frac{\pi}{4} \le \frac{\pi}{2} - \tan^{-1}x \le \frac{\pi}{2}$. Hence, $\frac{\pi}{4} \le \theta \le \frac{\pi}{2}$.