(a) Evaluate: (sin^3 60^°- tan 30^°)/(cos^2 45^°)

\[\begin{aligned} & \frac{\sin ^{3} 60^{\circ}-\tan 30^{\circ}}{\cos ^{2} 45^{\circ}}=\frac{\left(\dfrac{\sqrt{3}}{2}\right)^{3}-\dfrac{1}{\sqrt{3}}}{\left(\dfrac{1}{\sqrt{2}}\right)^{2}} \\ & =\frac{1}{4 \sqrt{3}} \text { or } \frac{\sqrt{3}}{12} \end{aligned} \]

\[\begin{aligned} & \tan (A+2 B)=\sqrt{3} \Rightarrow A+2 B=60^{\circ} \\ & \sin (2 A+B)=\frac{1}{\sqrt{2}} \Rightarrow 2 A+B=45^{\circ} \end{aligned} \]

On solving above equations, \(\displaystyle \mathrm{A}=10^{\circ}, \mathrm{B}=25^{\circ}\)