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MHTCET 17 Apr 2026 Shift 2 Paper

Question : 116 of 150

Marks: +1, -0

\cos ^{-1}\left(\cos \frac{4 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{4 \pi}{3}\right)=\ldots

Solution:

Concept:

Use the principal value ranges of inverse trigonometric functions.

Explanation:

For

\cos^{-1} x

, the principal range is

[0,\pi]

For

\sin^{-1} x

, the principal range is

\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

Now,

\frac{4\pi}{3}

is outside these principal ranges, so evaluate using known trigonometric values.

\cos \frac{4\pi}{3} = -\frac{1}{2}

, so

\cos^{-1}\left(\cos \frac{4\pi}{3}\right) = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}

\sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}

, so

\sin^{-1}\left(\sin \frac{4\pi}{3}\right) = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}

Adding them:

\frac{2\pi}{3} + \left(-\frac{\pi}{3}\right) = \frac{\pi}{3}

Answer:

\frac{\pi}{3}

, which is Option C.

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