Question:

Step 1: Determine the value of angle $A$. Given that $A$ is an acute angle and $\sin A = \dfrac{1}{2}$, it follows that $A = 30^\circ$. Step 2: Determine the sum of angles $A+B$. The conditions provided in the problem, in conjunction with the correct statements, imply that $A+B = 45^\circ$. Step 3: Verify the given statements based on $A+B = 45^\circ$. * **Statement 1:** $\sin^2(A+B) = \dfrac{1}{2}$ Substituting $A+B = 45^\circ$: $$ \sin^2(45^\circ) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} $$ This statement is correct. * **Statement 2:** $\tan\left(\dfrac{A+B}{2}\right) = \sqrt{2} - 1$ Substituting $A+B = 45^\circ$: $$ \tan\left(\frac{45^\circ}{2}\right) = \tan(22.5^\circ) $$ Using the half-angle identity $\tan\left(\frac{\theta}{2}\right) = \csc\theta - \cot\theta$: $$ \tan(22.5^\circ) = \csc(45^\circ) - \cot(45^\circ) = \sqrt{2} - 1 $$ This statement is correct. * **Statement 3:** $\cot\left(\dfrac{A+B}{3}\right) = 2 - \sqrt{3}$ Substituting $A+B = 45^\circ$: $$ \cot\left(\frac{45^\circ}{3}\right) = \cot(15^\circ) $$ We know that $\tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$. Rationalizing the denominator: $$ \tan(15^\circ) = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} $$ Therefore, $\cot(15^\circ) = \frac{1}{\tan(15^\circ)} = \frac{1}{2 - \sqrt{3}}$. Rationalizing the denominator: $$ \cot(15^\circ) = \frac{1(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3} $$ Since $2+\sqrt{3} \neq 2-\sqrt{3}$, this statement is incorrect. * **Statement 4:** $\cos(2A + 2B) = 0$ Substituting $A+B = 45^\circ$: $$ \cos(2(A+B)) = \cos(2 \times 45^\circ) = \cos(90^\circ) = 0 $$ This statement is correct.