Trigonometric Identities Question 180
Question: $ {{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta = $
[MP PET 1995, 2002; DCE 2005]
Options:
A) 0
B) -1
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3,{{\sin }^{2}}\theta ,{{\cos }^{2}}\theta $ $ ={{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{3}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1. $ Trick: Put $ \theta =0^{o}, $ we get the value of expression equal to 1. Again put $ \theta =45^{o}, $ the value remains 1, it means that the expression is independent of q and is equal to 1.
BETA
