Taking `A=30^(@)`, verify the following:
`tan2A=(2tanA)/(1-tan^(2)A)`

Answer

To verify the equation \( \tan(2A) = \frac{2\tan(A)}{1 - \tan^2(A)} \) for \( A = 30^\circ \), we will calculate both the left-hand side (LHS) and the right-hand side (RHS) and check if they are equal.

### Step 1: Calculate LHS
The left-hand side is given by:
\[ \text{LHS} = \tan(2A) = \tan(2 \times 30^\circ) = \tan(60^\circ) \]
From trigonometric values, we know:
\[ \tan(60^\circ) = \sqrt{3} \]

### Step 2: Calculate RHS
The right-hand side is given by:
\[ \text{RHS} = \frac{2\tan(A)}{1 - \tan^2(A)} = \frac{2\tan(30^\circ)}{1 - \tan^2(30^\circ)} \]
We know:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Now substituting this value into the RHS:
\[ \text{RHS} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} \]
Calculating \( \tan^2(30^\circ) \):
\[ \tan^2(30^\circ) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]
Now substituting this into the RHS:
\[ \text{RHS} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{\frac{2}{3}} \]
Now simplify the fraction:
\[ \text{RHS} = \frac{2 \cdot \frac{1}{\sqrt{3}} \cdot 3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Step 3: Compare LHS and RHS
Now we have:
\[ \text{LHS} = \sqrt{3} \quad \text{and} \quad \text{RHS} = \sqrt{3} \]
Since LHS = RHS, we have verified that:
\[ \tan(2A) = \frac{2\tan(A)}{1 - \tan^2(A)} \quad \text{is true for} \quad A = 30^\circ. \]