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Question : 101 of 150
Marks:
+1,
-0
Solution:
We have,
=π− and
=π− ⇒sin‌=sin(π−) and
sin‌=sin(π−) ⇒sin‌=sin‌ and
sin‌=sin‌ ⇒sin4()=sin4() and
sin4()=sin4() Now,
sin4()+sin4() +sin4()+sin4() =sin4()+sin4()+sin4() +sin4() =2sin4()+2sin4() =2[(sin2)2+(sin2())2] =2[()2 +()2]=2[+(1−cos‌)24] =[(1−)2+(1+)2] [∵cos‌= and
cos‌=] =[1+−+1++] =[2+]=[2+1]=
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