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Question : 110 of 160
Marks:
+1,
-0
Solution:
We have,
‌sin‌−1x+sin‌−1(1−x)=cos−1x‌⇒‌‌sin‌−1(1−x)=‌−sin‌−1x−sin‌−1x‌=‌−2sin‌−1x‌[∵cos−1x+sin‌−1x=‌]‌⇒sin‌−1(1−x)=sin‌−11−sin‌−12x√1−x2‌⇒sin‌−1(1−x)=sin‌−1(1√1−4x2(1−x2)−0)‌⇒‌‌(1−x)=√1−4x4+4x2=√(1−2x2)2‌⇒‌‌(1−x)=1−2x2‌⇒‌‌2x2−x=0‌⇒‌‌x(2x−1)=0‌⇒‌‌x=0,‌‌∴‌‌x∈{0,‌}‌
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