a) Ta có: \(A = {\cos ^2}{30^0} - {\sin ^2}{30^0}\)\( = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2}\) \( = \dfrac{3}{4} - \dfrac{1}{4} = \dfrac{1}{2}\)
\(B = \cos {60^0} + \sin {45^0}\)\( = \dfrac{1}{2} + \dfrac{{\sqrt 2 }}{2} = \dfrac{{1 + \sqrt 2 }}{2}\)
Vì \(\dfrac{{1 + \sqrt 2 }}{2} > \dfrac{1}{2}\) nên \(B > A\).
b) \(C = \dfrac{{2\tan {{30}^0}}}{{1 - {{\tan }^2}{{30}^0}}}\)\( = \dfrac{{2.\dfrac{1}{{\sqrt 3 }}}}{{1 - {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}}} = \dfrac{{\dfrac{2}{{\sqrt 3 }}}}{{1 - \dfrac{1}{3}}}\) \( = \dfrac{2}{{\sqrt 3 }}:\dfrac{2}{3} = \dfrac{2}{{\sqrt 3 }}.\dfrac{3}{2} = \sqrt 3 \)
\(D = ( - \tan {135^0}).tan{60^0}\)\( = \tan {45^0}.\tan {60^0} = 1.\sqrt 3 = \sqrt 3 \)
Vậy \(C = D\).
