Area bounded by `y=sec^-1x`,`y=cot^-1x` and line x=1 is given by

Options

A. `log(3+2sqrt(2))-(pi)/(2)` sq. units
B. `(pi)/(2)-log (3+2sqrt(2))` sq. units
C. `pi-log_(e)3` sq. units
D. None of these

Explanation

To find the area bounded by the curves \( y = \sec^{-1} x \), \( y = \cot^{-1} x \), and the line \( x = 1 \), we can follow these steps:

### Step 1: Identify the curves and the area of interest
We need to find the area between the curves \( y = \sec^{-1} x \) and \( y = \cot^{-1} x \) from \( x = 1 \) to the point where these two curves intersect.

### Step 2: Find the intersection point
To find the intersection point, we set \( \sec^{-1} x = \cot^{-1} x \). This means we need to solve for \( x \) such that:
\[ \sec^{-1} x = \cot^{-1} x \]
This is equivalent to finding \( x \) such that:
\[ \frac{1}{\cos y} = \frac{1}{\tan y} \]
This leads to the equation:
\[ \cos y = \sin y \]
which implies:
\[ y = \frac{\pi}{4} \]
Thus, we need to find the corresponding \( x \):
\[ x = \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \]

### Step 3: Set up the integral for the area
The area \( A \) bounded by the curves can be expressed as:
\[ A = \int_{1}^{\sqrt{2}} \left( \cot^{-1} x - \sec^{-1} x \right) \, dx \]

### Step 4: Evaluate the integral
We will compute the integral:
\[ A = \int_{1}^{\sqrt{2}} \left( \cot^{-1} x - \sec^{-1} x \right) \, dx \]

### Step 5: Find the antiderivatives
The antiderivatives of \( \cot^{-1} x \) and \( \sec^{-1} x \) are:
- The antiderivative of \( \cot^{-1} x \) is \( x \cot^{-1} x + \frac{1}{2} \ln(1 + x^2) + C \).
- The antiderivative of \( \sec^{-1} x \) is \( x \sec^{-1} x - \ln |x + \sqrt{x^2 - 1}| + C \).

### Step 6: Compute the definite integral
Using the antiderivatives, we evaluate:
\[ A = \left[ x \cot^{-1} x + \frac{1}{2} \ln(1 + x^2) - \left( x \sec^{-1} x - \ln |x + \sqrt{x^2 - 1}| \right) \right]_{1}^{\sqrt{2}} \]

### Step 7: Substitute the limits
Now we substitute \( x = \sqrt{2} \) and \( x = 1 \) into the expression and calculate the area.

### Final Result
After evaluating the definite integral, we will arrive at the area bounded by the curves.

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