What is sin `[cot^(-1){cos(tan^(-1)x)]` where x `gt` 0, equal to ?

Options

A. `sqrt(((x^(2)+1))/((x^(2)+2)))`
B. `sqrt(((x^(2)+2))/((x^(2)+1)))`
C. `((x^(2)+1))/((x^(2)+2))`
D. `((x^(2)+2))/((x^(2)+1))`

Explanation

To solve the problem \( \sin \left[ \cot^{-1} \left( \cos \left( \tan^{-1} x \right) \right) \right] \), where \( x > 0 \), we will follow these steps:

### Step 1: Let \( \alpha = \tan^{-1} x \)
From this definition, we know that:
\[ \tan \alpha = x \]

### Step 2: Find \( \sec \alpha \)
Using the identity \( \sec^2 \alpha = 1 + \tan^2 \alpha \):
\[ \sec^2 \alpha = 1 + x^2 \implies \sec \alpha = \sqrt{1 + x^2} \]

### Step 3: Find \( \cos \alpha \)
Since \( \cos \alpha = \frac{1}{\sec \alpha} \):
\[ \cos \alpha = \frac{1}{\sqrt{1 + x^2}} \]

### Step 4: Let \( \beta = \cot^{-1} \left( \cos \alpha \right) \)
Thus, we have:
\[ \cot \beta = \cos \alpha = \frac{1}{\sqrt{1 + x^2}} \]

### Step 5: Find \( \sin \beta \)
Using the identity \( \sin^2 \beta + \cos^2 \beta = 1 \), we can express \( \sin \beta \):
\[ \sin \beta = \frac{1}{\sqrt{1 + \cot^2 \beta}} = \frac{1}{\sqrt{1 + \left( \frac{1}{\sqrt{1 + x^2}} \right)^2}} \]

### Step 6: Simplify \( \sin \beta \)
Calculating \( \cot^2 \beta \):
\[ \cot^2 \beta = \frac{1}{1 + x^2} \]
Thus:
\[ \sin \beta = \frac{1}{\sqrt{1 + \frac{1}{1 + x^2}}} = \frac{1}{\sqrt{\frac{1 + x^2 + 1}{1 + x^2}}} = \frac{1}{\sqrt{\frac{2 + x^2}{1 + x^2}}} \]

### Step 7: Final expression for \( \sin \left[ \cot^{-1} \left( \cos \left( \tan^{-1} x \right) \right) \right] \)
Now we can express \( \sin \beta \):
\[ \sin \beta = \frac{\sqrt{1 + x^2}}{\sqrt{2 + x^2}} \]

Thus, the final answer is:
\[ \sin \left[ \cot^{-1} \left( \cos \left( \tan^{-1} x \right) \right) \right] = \frac{\sqrt{1 + x^2}}{\sqrt{2 + x^2}} \]