Inverse Trigonometric Functions
NCERT Exemplar Class 12
CBSE
Grade 12
Question:
The value of $\sin^{-1}\left(\sin \dfrac{2\pi}{3}\right)$ is:
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{2\pi}{3}$
(c) $-\dfrac{\pi}{3}$
(d) $\dfrac{4\pi}{3}$
Step-by-Step Solution
\sin(2\pi/3) = \sin(\pi/3) \Rightarrow \sin^{-1}(\sin(\pi/3)) = \pi/3. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Selecting $\pi/3$: 1.0 Mark
Correct Answer: $\dfrac{\pi}{3}$
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