Inverse Trigonometric Functions
NCERT Exemplar Class 12
CBSE
Grade 12
Question:
Find the value of $\sin\left(\dfrac{\pi}{3} - \sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)\right)$.
Step-by-Step Solution
\sin^{-1}(-\sqrt{3}/2) = -\pi/3. [1.0 Mark]
\sin(\pi/3 + \pi/3) = \sin(2\pi/3) = \sqrt{3}/2. [1.0 Mark]
---
🎯 Official CBSE Marking Scheme:
Evaluating $\sin^{-1}(-\sqrt{3}/2) = -\pi/3$: 1.0 Mark
Evaluating final value $= \sqrt{3}/2$: 1.0 Mark
Correct Answer:
Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.