Verify that $\displaystyle sin 2A = \frac{2 tan A}{1 + tan^2A} for A= 30^\circ$

Verify that $\displaystyle sin 2A = \frac{2 tan A}{1 + tan^2A} for A= 30^\circ$

We need to verify that:

$ \displaystyle \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} $

for $ \displaystyle A = 30^\circ $.

### Step 1: Compute Left-Hand Side (LHS)
The LHS is:

$ \sin 2A = \sin (2 \times 30^\circ) = \sin 60^\circ $

Using the known trigonometric value:

$ \sin 60^\circ = \frac{\sqrt{3}}{2} $

### Step 2: Compute Right-Hand Side (RHS)
The RHS is:

$ \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} $

Using the known values:

$ \tan 30^\circ = \frac{1}{\sqrt{3}} $, so:

$ \tan^2 30^\circ = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} $

Substituting these values into the RHS:

$ \displaystyle \frac{2 \times \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} $

Simplify the denominator:

$ \displaystyle 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} $

Now the expression becomes:

$ \displaystyle \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} $

Multiplying by the reciprocal:

$ \displaystyle \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} $

Rationalizing the denominator:

$ \displaystyle \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2} $

### Step 3: Compare LHS and RHS
Since both sides are equal:

$ \displaystyle \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} $

Thus, the identity is verified for $ A = 30^\circ $.

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