Let `f(x)=(sin^(-1)x)/(cos^(-1)x)+(cos^(-1)x)/(tan^(-1)x)+(tan^(-1)x)/(sec^(-1)x)+(sec^(-1)x)/(cosec^(-1)x)+(cosec^(-1)x)/(cot^(-1)x)+(cot^(-1)x)/(sin^(-1)x)`. Then which of the following statements holds good?

Options

A. a. Minimum value of f(x) is 6.
B. b. `f(x)` is a continous function.
C. c. `f(-1)=(-107)/(12)`
D. d. `f(x)` is non-derivable at `x=-1`

Explanation

To solve the problem, we need to analyze the function given:

\[ f(x) = \frac{\sin^{-1} x}{\cos^{-1} x} + \frac{\cos^{-1} x}{\tan^{-1} x} + \frac{\tan^{-1} x}{\sec^{-1} x} + \frac{\sec^{-1} x}{\csc^{-1} x} + \frac{\csc^{-1} x}{\cot^{-1} x} + \frac{\cot^{-1} x}{\sin^{-1} x} \]

### Step 1: Understand the Domains of the Inverse Functions
The domains of the inverse trigonometric functions are as follows:
- \(\sin^{-1} x\) is defined for \(x \in [-1, 1]\)
- \(\cos^{-1} x\) is defined for \(x \in [-1, 1]\)
- \(\tan^{-1} x\) is defined for all \(x \in (-\infty, \infty)\)
- \(\sec^{-1} x\) is defined for \(x \in (-\infty, -1] \cup [1, \infty)\)
- \(\csc^{-1} x\) is defined for \(x \in (-\infty, -1] \cup [1, \infty)\)
- \(\cot^{-1} x\) is defined for all \(x \in (-\infty, \infty)\)

### Step 2: Determine the Common Domain
The common domain for the function \(f(x)\) is determined by the most restrictive domains of the components:
- The common domain is \(x \in [-1, 1]\) because both \(\sin^{-1} x\) and \(\cos^{-1} x\) restrict \(x\) to this interval.

### Step 3: Evaluate \(f(x)\) at the Endpoints
Next, we evaluate \(f(x)\) at the endpoints of the domain, \(x = -1\) and \(x = 1\).

1. **At \(x = 1\)**:
\[ f(1) = \frac{\sin^{-1}(1)}{\cos^{-1}(1)} + \frac{\cos^{-1}(1)}{\tan^{-1}(1)} + \frac{\tan^{-1}(1)}{\sec^{-1}(1)} + \frac{\sec^{-1}(1)}{\csc^{-1}(1)} + \frac{\csc^{-1}(1)}{\cot^{-1}(1)} + \frac{\cot^{-1}(1)}{\sin^{-1}(1)} \]
- \(\sin^{-1}(1) = \frac{\pi}{2}\)
- \(\cos^{-1}(1) = 0\) (undefined division)
- Therefore, \(f(1)\) is undefined.

2. **At \(x = -1\)**:
\[ f(-1) = \frac{\sin^{-1}(-1)}{\cos^{-1}(-1)} + \frac{\cos^{-1}(-1)}{\tan^{-1}(-1)} + \frac{\tan^{-1}(-1)}{\sec^{-1}(-1)} + \frac{\sec^{-1}(-1)}{\csc^{-1}(-1)} + \frac{\csc^{-1}(-1)}{\cot^{-1}(-1)} + \frac{\cot^{-1}(-1)}{\sin^{-1}(-1)} \]
- \(\sin^{-1}(-1) = -\frac{\pi}{2}\)
- \(\cos^{-1}(-1) = \pi\)
- \(\tan^{-1}(-1) = -\frac{\pi}{4}\)
- \(\sec^{-1}(-1) = \pi\)
- \(\csc^{-1}(-1) = -\frac{\pi}{2}\)
- \(\cot^{-1}(-1) = \frac{3\pi}{4}\)

Evaluating \(f(-1)\) gives:
\[ f(-1) = \frac{-\frac{\pi}{2}}{\pi} + \frac{\pi}{-\frac{\pi}{4}} + \frac{-\frac{\pi}{4}}{\pi} + \frac{\pi}{-\frac{\pi}{2}} + \frac{-\frac{\pi}{2}}{\frac{3\pi}{4}} + \frac{\frac{3\pi}{4}}{-\frac{\pi}{2}} \]
Simplifying each term:
- First term: \(-\frac{1}{2}\)
- Second term: \(-4\)
- Third term: \(-\frac{1}{4}\)
- Fourth term: \(-2\)
- Fifth term: \(-\frac{3}{2}\)
- Sixth term: \(-\frac{3}{2}\)

Adding these values gives:
\[ f(-1) = -\frac{1}{2} - 4 - \frac{1}{4} - 2 - \frac{3}{2} - \frac{3}{2} = -\frac{1}{2} - 4 - \frac{1}{4} - 2 - 3 = -\frac{39}{4} \]

### Step 4: Conclusion
Since \(f(1)\) is undefined and \(f(-1)\) is a finite value, we can conclude that \(f(x)\) is not continuous over the interval \([-1, 1]\) and is non-derivable at \(x = 1\).

### Final Answer
The correct statement is that \(f(x)\) is non-derivable at \(x = 1\) and the minimum value of \(f(x)\) is not defined.