If `A = 2 tan^(-1) (2 sqrt2 -1) and B = 3 sin^(-1) ((1)/(3)) + sin^(-1) ((3)/(5))`, then which is greater ?

Answer

To solve the problem, we need to evaluate \( A \) and \( B \) separately and then compare them.

### Step 1: Evaluate \( A = 2 \tan^{-1}(2\sqrt{2} - 1) \)

1. Substitute the value of \( \sqrt{2} \) (approximately 1.414):
\[ A = 2 \tan^{-1}(2 \times 1.414 - 1) = 2 \tan^{-1}(2.828 - 1) = 2 \tan^{-1}(1.828) \]

2. We know that \( 1.828 \) is greater than \( \sqrt{3} \) (approximately 1.732). Hence, we can say:
\[ \tan^{-1}(1.828) > \tan^{-1}(\sqrt{3}) \]

3. Since \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \), we have:
\[ \tan^{-1}(1.828) > \frac{\pi}{3} \]

4. Therefore:
\[ A = 2 \tan^{-1}(1.828) > 2 \times \frac{\pi}{3} = \frac{2\pi}{3} \]

### Step 2: Evaluate \( B = 3 \sin^{-1}\left(\frac{1}{3}\right) + \sin^{-1}\left(\frac{3}{5}\right) \)

1. First, evaluate \( 3 \sin^{-1}\left(\frac{1}{3}\right) \):
- Using the identity \( 3 \sin^{-1}(x) = \sin^{-1}(3x - 4x^3) \):
\[ 3 \sin^{-1}\left(\frac{1}{3}\right) = \sin^{-1}\left(3 \cdot \frac{1}{3} - 4 \cdot \left(\frac{1}{3}\right)^3\right) = \sin^{-1}\left(1 - \frac{4}{27}\right) = \sin^{-1}\left(\frac{23}{27}\right) \]

2. Now, add \( \sin^{-1}\left(\frac{3}{5}\right) \):
\[ B = \sin^{-1}\left(\frac{23}{27}\right) + \sin^{-1}\left(\frac{3}{5}\right) \]

3. Since both \( \sin^{-1}\left(\frac{23}{27}\right) \) and \( \sin^{-1}\left(\frac{3}{5}\right) \) are positive angles, we can estimate their values:
- \( \sin^{-1}\left(\frac{23}{27}\right) \) is approximately \( 0.851 \) radians.
- \( \sin^{-1}\left(\frac{3}{5}\right) \) is approximately \( 0.6435 \) radians.

4. Thus:
\[ B \approx 0.851 + 0.6435 = 1.4945 \text{ radians} \]

### Step 3: Compare \( A \) and \( B \)

1. We found that:
\[ A > \frac{2\pi}{3} \approx 2.094 \text{ radians} \]
and
\[ B \approx 1.4945 \text{ radians} \]

2. Therefore, we conclude that:
\[ A > B \]

### Final Answer:
\( A \) is greater than \( B \).