Question

Verify that \(\sin 2A = \frac{2 \tan A}{1 + \tan^2 A}\) for \(A = 30^\circ\).

Answer 1

Verification of the Identity

Verify the identity: \[ \sin 2A = \frac{2 \tan A}{1 + \tan^2 A} \quad \text{for} \quad A = 30^\circ \]

Left Hand Side (LHS):

\[ \sin 2A = \sin(2 \times 30^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

Right Hand Side (RHS):

We know: \[ \tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \tan^2 30^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

Substitute into the RHS expression:

\[ \frac{2 \tan A}{1 + \tan^2 A} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} \]

Rationalizing the denominator:

\[ \frac{3}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{2 \cdot 3} = \frac{\sqrt{3}}{2} \]

Conclusion:

\[ \text{LHS} = \frac{\sqrt{3}}{2}, \quad \text{RHS} = \frac{\sqrt{3}}{2} \Rightarrow \text{LHS = RHS} \]