• # question_answer If ${{\cos }^{6}}\alpha +{{\sin }^{6}}\alpha +K\,{{\sin }^{2}}2\alpha =1,$ then K = A) $\frac{4}{3}$ B) $\frac{3}{4}$ C) $\frac{1}{2}$ D) 2

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. $K=\frac{3}{4}$.