If `x_1=cos^(-1)(3/5)+cos^(-1)((2sqrt2)/3)and x_2=sin^(-1)(3/5)+sin^(-1)((2sqrt2)/3)`, then

Options

A. `x_1 lt x_2`
B. `x_1=x_2`
C. `x_1 gt x_2`
D. Can't be determined

Answer

To solve the problem, we need to find the relationship between \( x_1 \) and \( x_2 \), where:

\[ x_1 = \cos^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{2\sqrt{2}}{3}\right) \]
\[ x_2 = \sin^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \]

### Step 1: Calculate \( x_1 \)

1. **Identify the angles**:
- Let \( \theta = \cos^{-1}\left(\frac{3}{5}\right) \) and \( \alpha = \cos^{-1}\left(\frac{2\sqrt{2}}{3}\right) \).

2. **Find the sides of the triangle**:
- For \( \theta \):
- Hypotenuse = 5, Base = 3
- Perpendicular = \( \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \)
- For \( \alpha \):
- Hypotenuse = 3, Base = \( 2\sqrt{2} \)
- Perpendicular = \( \sqrt{3^2 - (2\sqrt{2})^2} = \sqrt{9 - 8} = \sqrt{1} = 1 \)

3. **Calculate \( \tan \theta \) and \( \tan \alpha \)**:
- \( \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{4}{3} \)
- \( \tan \alpha = \frac{1}{2\sqrt{2}} \)

4. **Express \( x_1 \) in terms of \( \tan^{-1} \)**:
\[ x_1 = \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{2\sqrt{2}}\right) \]

5. **Use the formula for \( \tan^{-1}(a) + \tan^{-1}(b) \)**:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
Here, \( a = \frac{4}{3} \) and \( b = \frac{1}{2\sqrt{2}} \).

6. **Calculate \( ab \)**:
\[ ab = \frac{4}{3} \cdot \frac{1}{2\sqrt{2}} = \frac{4}{6\sqrt{2}} = \frac{2}{3\sqrt{2}} \]

7. **Calculate \( x_1 \)**:
\[ x_1 = \tan^{-1}\left(\frac{\frac{4}{3} + \frac{1}{2\sqrt{2}}}{1 - \frac{2}{3\sqrt{2}}}\right) \]

### Step 2: Calculate \( x_2 \)

1. **Identify the angles**:
- Let \( \theta' = \sin^{-1}\left(\frac{3}{5}\right) \) and \( \alpha' = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \).

2. **Find the sides of the triangle**:
- For \( \theta' \):
- Hypotenuse = 5, Perpendicular = 3
- Base = \( \sqrt{5^2 - 3^2} = \sqrt{16} = 4 \)
- For \( \alpha' \):
- Hypotenuse = 3, Perpendicular = \( 2\sqrt{2} \)
- Base = \( \sqrt{3^2 - (2\sqrt{2})^2} = \sqrt{1} = 1 \)

3. **Calculate \( \tan \theta' \) and \( \tan \alpha' \)**:
- \( \tan \theta' = \frac{3}{4} \)
- \( \tan \alpha' = \frac{2\sqrt{2}}{1} \)

4. **Express \( x_2 \) in terms of \( \tan^{-1} \)**:
\[ x_2 = \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(2\sqrt{2}\right) \]

5. **Use the formula for \( \tan^{-1}(a) + \tan^{-1}(b) \)**:
\[ x_2 = \tan^{-1}\left(\frac{\frac{3}{4} + 2\sqrt{2}}{1 - \frac{3}{4} \cdot 2\sqrt{2}}\right) \]

### Step 3: Compare \( x_1 \) and \( x_2 \)

1. **From the calculations**:
- \( x_1 \) is expressed as a positive angle.
- \( x_2 \) involves a negative term when simplified.

2. **Conclusion**:
- Since \( x_1 \) is positive and \( x_2 \) can be expressed in terms of a negative angle, we conclude that \( x_1 > x_2 \).

### Final Result
Thus, we find that:

\[ x_1 > x_2 \]