a) \(A=cos^230^o-sin^230^o=\left(\dfrac{\sqrt{3}}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2}\); \(B=cos60^o+sin45^o=\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}\). Vì vậy \(A< B\). b) \(C=\dfrac{2tan30^o}{1-tan^230^o}=\dfrac{2\dfrac{\sqrt{3}}{2}}{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}=\sqrt{3}\). \(D=\left(-tan135^o\right)tan60^o=-\left(-1\right).\sqrt{3}=\sqrt{3}\). Vậy \(C=D\).