If `tanA=sqrt(2)-1`, then `sin2A` = ?

Options

A. `(1)/(sqrt(2))`
B. `sqrt(2)`
C. 2
D. `2sqrt(2)`

Answer

To solve the problem where \( \tan A = \sqrt{2} - 1 \) and we need to find \( \sin 2A \), we can follow these steps:

### Step 1: Use the double angle formula for tangent
The formula for \( \tan 2A \) is given by:
\[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \]
Substituting \( \tan A = \sqrt{2} - 1 \) into the formula, we get:
\[ \tan 2A = \frac{2(\sqrt{2} - 1)}{1 - (\sqrt{2} - 1)^2} \]

### Step 2: Calculate \( \tan^2 A \)
First, we need to calculate \( \tan^2 A \):
\[ \tan^2 A = (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \]

### Step 3: Substitute \( \tan^2 A \) into the formula
Now, substitute \( \tan^2 A \) back into the formula for \( \tan 2A \):
\[ \tan 2A = \frac{2(\sqrt{2} - 1)}{1 - (3 - 2\sqrt{2})} \]
This simplifies to:
\[ \tan 2A = \frac{2(\sqrt{2} - 1)}{1 - 3 + 2\sqrt{2}} = \frac{2(\sqrt{2} - 1)}{2\sqrt{2} - 2} \]

### Step 4: Simplify the expression
We can simplify the expression further:
\[ \tan 2A = \frac{2(\sqrt{2} - 1)}{2(\sqrt{2} - 1)} = 1 \]

### Step 5: Find the angle \( 2A \)
Since \( \tan 2A = 1 \), we know:
\[ 2A = 45^\circ \quad \text{(or any angle where tangent equals 1)} \]

### Step 6: Find \( \sin 2A \)
Now that we have \( 2A = 45^\circ \), we can find \( \sin 2A \):
\[ \sin 2A = \sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Final Answer
Thus, the value of \( \sin 2A \) is:
\[ \sin 2A = \frac{\sqrt{2}}{2} \]

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