Trigonometric Identities Question 68
Question: If $ {{\cos }^{6}}\alpha +{{\sin }^{6}}\alpha +K,{{\sin }^{2}}2\alpha =1, $ then K =
Options:
A) $ \frac{4}{3} $
B) $ \frac{3}{4} $
C) $ \frac{1}{2} $
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ {{\cos }^{6}}\alpha +{{\sin }^{6}}\alpha +K{{\sin }^{2}}2\alpha =1 $ using formula $ a^{3}+b^{3}={{(a+b)}^{3}}-3ab(a+b) $ and on solving, we get the required result i.e. $ K=\frac{3}{4} $ .
BETA
