`|{:(x,e^(x-1),(x-1)^(3)),(x-lnx,cos(x-1),(x-1)^(2)),(tanx,sin^(2)x,cos^(2)x):}|=a_(0)+a_(1)(x-1)+a_(2)(x-1)^(2)cdots`

The value of `cos^(-1) (a_(1))` is:

Options

A. (a) `0`
B. (b) `(pi)/(4)`
C. (c) `(pi)/(2)`
D. (d) `pi`

Answer

To solve the given determinant problem, we will follow these steps:

### Step 1: Define the Determinant
We have the determinant defined as:
\[ \Delta(x) = \begin{vmatrix} x & e^{x-1} & (x-1)^3 \\ x - \ln x & \cos(x-1) & (x-1)^2 \\ \tan x & \sin^2 x & \cos^2 x \end{vmatrix} \]

### Step 2: Expand the Determinant
We need to express this determinant as a polynomial in terms of \( (x-1) \):
\[ \Delta(x) = a_0 + a_1(x-1) + a_2(x-1)^2 + \cdots \]

### Step 3: Find the Derivative
To find \( a_1 \), we differentiate \( \Delta(x) \) with respect to \( x \):
\[ \Delta'(x) = \frac{d}{dx} \Delta(x) \]
When we differentiate, the constant term \( a_0 \) will vanish, and we will have:
\[ \Delta'(x) = a_1 + 2a_2(x-1) + 3a_3(x-1)^2 + \cdots \]

### Step 4: Evaluate the Derivative at \( x = 1 \)
We evaluate \( \Delta'(x) \) at \( x = 1 \):
\[ \Delta'(1) = a_1 \]

### Step 5: Calculate \( \Delta'(x) \)
Now we need to compute \( \Delta'(x) \) explicitly. For the determinant:
\[ \Delta'(x) = \begin{vmatrix} 1 & e^{x-1} & 3(x-1)^2 \\ 1 - \frac{1}{x} & -\sin(x-1) & 2(x-1) \\ \sec^2 x & 2\sin x \cos x & -2\sin x \cos x \end{vmatrix} \]

### Step 6: Evaluate \( \Delta'(1) \)
Now we evaluate \( \Delta'(1) \):
\[ \Delta'(1) = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{vmatrix} \]
Since one row is entirely zeros, the determinant evaluates to 0:
\[ \Delta'(1) = 0 \]
Thus, \( a_1 = 0 \).

### Step 7: Find \( \cos^{-1}(a_1) \)
Now we need to find \( \cos^{-1}(a_1) \):
\[ \cos^{-1}(0) = \frac{\pi}{2} \]

### Conclusion
The value of \( \cos^{-1}(a_1) \) is \( \frac{\pi}{2} \).