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Question : 160 of 180
Marks:
+1,
-0
Solution:
We have,
‌sin‌(2sin‌−1√‌)‌‌=sin‌{sin‌−1(2√‌√1−(√‌)2)}‌[∵2sin‌−1x.=sin‌−12x√1−x2] ‌=sin‌{sin‌−1(2√‌√1−‌)}‌=sin‌{sin‌−1(2√‌×√‌)}=sin‌{sin‌−1‌}‌=‌‌‌[∵sin‌(sin‌−1x)=x]
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