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Question : 21 of 160
Marks:
+1,
-0
Solution:
We have,
‌√sin‌4x+4cos2x−√cos4x+4sin‌2x‌=√sin‌4x+4−4sin‌2x−√cos4x+4−4cos2x
‌=√(2−sin‌2x)2−√(2−2cos2x)2‌=2−sin‌2x−2+cos2x=cos2x−sin‌2x‌=cos‌2‌x
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