Question 2

We will use trigonometric identities and known values to simplify expressions.

Step 1 — Evaluate part (i)

We recognize a trigonometric identity. Double Angle Identity: $\sin 2A = \dfrac{2\tan A}{1 + \tan^2 A}$ — this converts a ratio expression involving $\tan A$ directly into $\sin$ of double the angle.

The expression matches $\sin 2A = \frac{2 \tan A}{1 + \tan^2 A}$. Here, $A$ is $\mathbf{30^\circ}$. We substitute $A = \mathbf{30^\circ}$ into the identity.

$\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \sin (2 \times 30^\circ)$

$= \sin 60^\circ$

$\boxed{\sin 60^\circ}$

Step 2 — Evaluate part (ii)

We recognize a trigonometric identity. Double Angle Identity: $\cos 2A = \dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ — converts a tan-based expression into $\cos$ of double the angle.

The expression matches $\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A}$. Here, $A$ is $\mathbf{45^\circ}$. We substitute $A = \mathbf{45^\circ}$ into the identity.

$\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \cos (2 \times 45^\circ)$

$= \cos 90^\circ$

$= 0$

$\boxed{0}$

Step 3 — Evaluate part (iii)

We need to find $A$ where $\sin 2A = 2 \sin A$. Let's check the given options. Consider option (A) where $A = \mathbf{0^\circ}$. First, calculate the left side (LHS).

$\text{LHS} = \sin (2 \times 0^\circ)$

$= \sin 0^\circ$

$= 0$ Next, calculate the right side (RHS).

$\text{RHS} = 2 \sin 0^\circ$

$= 2 \times 0$

$= 0$ Both sides are equal to $\mathbf{0}$. So, $A = \mathbf{0^\circ}$ is the correct value.

$\boxed{A = 0^\circ}$

Step 4 — Evaluate part (iv)

We recognize a trigonometric identity. The expression matches $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$. Here, $A$ is $\mathbf{30^\circ}$. We substitute $A = \mathbf{30^\circ}$ into the identity.

$\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = \tan (2 \times 30^\circ)$

$= \tan 60^\circ$

$\boxed{\tan 60^\circ}$

Answer

(i) (A) $\sin 60^\circ$ (ii) (D) $0$ (iii) (A) $0^\circ$ (iv) (C) $\tan 60^\circ$