The area bounded by the curve `y=|cos^(-1)(sinx)|-|sin^(-1)(cosx)|` and axis from `(3pi)/(2)lex le 2pi`

Options

A. `pi^(2)` sq. units
B. `pi^(2)//4` sq. units
C. `pi^(2)//2` sq. units
D. none of these

Explanation

To find the area bounded by the curve \( y = |\cos^{-1}(\sin x)| - |\sin^{-1}(\cos x)| \) and the axes from \( x = \frac{3\pi}{2} \) to \( x = 2\pi \), we will follow these steps:

### Step 1: Simplify the expression for \( y \)

We start with the expression:
\[ y = |\cos^{-1}(\sin x)| - |\sin^{-1}(\cos x)| \]

For \( x \) in the interval \( \left[\frac{3\pi}{2}, 2\pi\right] \):
- \( \sin x \) will take values from \( -1 \) to \( 0 \).
- \( \cos x \) will take values from \( 0 \) to \( 1 \).

Thus, we can simplify:
\[ \sin^{-1}(\cos x) = \sin^{-1}(\cos x) = \sin^{-1}(0) = 0 \quad \text{at } x = \frac{3\pi}{2} \]
\[ \sin^{-1}(\cos x) = \sin^{-1}(1) = \frac{\pi}{2} \quad \text{at } x = 2\pi \]

Using the properties of inverse trigonometric functions:
- \( \cos^{-1}(\sin x) = \frac{\pi}{2} - x \) when \( x \) is in the range \( \left[\frac{3\pi}{2}, 2\pi\right] \).

Thus, we have:
\[ y = \left|\frac{\pi}{2} - x\right| - \left|\sin^{-1}(\cos x)\right| \]

### Step 2: Evaluate the expressions

Now, we need to evaluate:
\[ y = \left|\frac{\pi}{2} - x\right| - \left|\frac{\pi}{2} - x\right| = 0 \]

This means that within the interval \( \left[\frac{3\pi}{2}, 2\pi\right] \), the value of \( y \) is consistently zero.

### Step 3: Determine the area

Since \( y = 0 \) for all \( x \) in the interval \( \left[\frac{3\pi}{2}, 2\pi\right] \), the area under the curve from \( x = \frac{3\pi}{2} \) to \( x = 2\pi \) is simply:
\[ \text{Area} = \int_{\frac{3\pi}{2}}^{2\pi} 0 \, dx = 0 \]

### Conclusion

Thus, the area bounded by the curve and the axes from \( x = \frac{3\pi}{2} \) to \( x = 2\pi \) is:
\[ \text{Area} = 0 \text{ square units} \]