Test Index

CBSE Class 12 Math 2023 Delhi Set 1 Solved Paper

© examsnet.com
Question : 2 of 38
Marks: +1, -0
sin⁡[π3+sin⁡−1(12)]\sin \left[ \frac{\pi}{3} + \sin^{-1} \left( \frac{1}{2} \right) \right] is equal to:
Solution:     👈: Video Solution
 Explanation: sin⁡[π3+sin⁡−1(12)]\text{ Explanation: } \sin \left[ \frac{\pi}{3} + \sin^{-1} \left( \frac{1}{2} \right) \right]
sin⁡(π3+π6)=sin⁡π2=1\sin \left( \frac{\pi}{3} + \frac{\pi}{6} \right) = \sin \frac{\pi}{2} = 1
© examsnet.com
Go to Question:
Prev Question
Next Question