sin^-1(cos x) = (π)/2 - x is valid for :

The correct answer is Option (2) → \(0 \le x \le \pi\)

Since \(\sin^{-1}y + \cos^{-1}y = \frac{\pi}{2}\) for every y in [−1, 1], with y = cos x:

    \( \displaystyle \sin^{-1}(\cos x) = \frac{\pi}{2} - \cos^{-1}(\cos x) \)

So the given result holds exactly when

    \(\cos^{-1}(\cos x) = x\)

The principal value branch of \(\cos^{-1}\) is [0, π], so \(\cos^{-1}(\cos x) = x\) only for \(0 \le x \le \pi\).

For example, x = \(-\frac{\pi}{4}\) lies in the intervals of the other three options, but there

    \( \displaystyle \sin^{-1}\left(\cos\left(-\frac{\pi}{4}\right)\right) = \sin^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4} \)

    \( \displaystyle \frac{\pi}{2} - \left(-\frac{\pi}{4}\right) = \frac{3\pi}{4} \ne \frac{\pi}{4} \)

Note: This topic was not in the CUET 2026 syllabus.