Q66 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
$\cos^{-1}\left(\cos\dfrac{4\pi}{3}\right) + \sin^{-1}\left(\sin\dfrac{4\pi}{3}\right) = \ldots$
- A$\dfrac{4\pi}{3}$
- B$\dfrac{8\pi}{3}$
- Correct answerC$\dfrac{\pi}{3}$
- D$\dfrac{3\pi}{2}$
Solution
The principal value branch of $\cos^{-1}x$ is $[0, \pi]$.
$\cos^{-1}\left(\cos\dfrac{4\pi}{3}\right) = \cos^{-1}\left(\cos\left(2\pi - \dfrac{2\pi}{3}\right)\right) = \cos^{-1}\left(\cos\dfrac{2\pi}{3}\right) = \dfrac{2\pi}{3}$
The principal value branch of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
$\sin^{-1}\left(\sin\dfrac{4\pi}{3}\right) = \sin^{-1}\left(\sin\left(\pi + \dfrac{\pi}{3}\right)\right) = \sin^{-1}\left(-\sin\dfrac{\pi}{3}\right) = \sin^{-1}\left(\sin\left(-\dfrac{\pi}{3}\right)\right) = -\dfrac{\pi}{3}$
Adding the two values:
$\dfrac{2\pi}{3} - \dfrac{\pi}{3} = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
$\cos^{-1}\left(\cos\dfrac{4\pi}{3}\right) = \cos^{-1}\left(\cos\left(2\pi - \dfrac{2\pi}{3}\right)\right) = \cos^{-1}\left(\cos\dfrac{2\pi}{3}\right) = \dfrac{2\pi}{3}$
The principal value branch of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
$\sin^{-1}\left(\sin\dfrac{4\pi}{3}\right) = \sin^{-1}\left(\sin\left(\pi + \dfrac{\pi}{3}\right)\right) = \sin^{-1}\left(-\sin\dfrac{\pi}{3}\right) = \sin^{-1}\left(\sin\left(-\dfrac{\pi}{3}\right)\right) = -\dfrac{\pi}{3}$
Adding the two values:
$\dfrac{2\pi}{3} - \dfrac{\pi}{3} = \dfrac{\pi}{3}$
Answer: $\dfrac{\pi}{3}$
Hence, the correct answer is option: 'C'.