The value of `sin[cot^(-1){cos(tan^(-1) x)}]` is

Options

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

Answer

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

### Step 1: Define the angle
Let \( \theta = \tan^{-1}(x) \). This means that \( \tan(\theta) = x \).

### Step 2: Construct a right triangle
From the definition of tangent, we can create a right triangle where:
- The opposite side (perpendicular) is \( x \)
- The adjacent side (base) is \( 1 \)

Using the Pythagorean theorem, the hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{x^2 + 1} \]

### Step 3: Find \( \cos(\theta) \)
Now, we can find \( \cos(\theta) \):
\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 4: Substitute into the cotangent function
Next, we need to find \( \cot^{-1}(\cos(\theta)) \):
\[ \cot^{-1}\left(\cos(\theta)\right) = \cot^{-1}\left(\frac{1}{\sqrt{x^2 + 1}}\right) \]

Let \( \phi = \cot^{-1}\left(\frac{1}{\sqrt{x^2 + 1}}\right) \). This means:
\[ \cot(\phi) = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 5: Construct another right triangle for \( \phi \)
From \( \cot(\phi) \), we can create another right triangle where:
- The adjacent side is \( 1 \)
- The opposite side is \( \sqrt{x^2 + 1} \)

Using the Pythagorean theorem again, the hypotenuse \( h' \) can be calculated as:
\[ h' = \sqrt{1^2 + \left(\sqrt{x^2 + 1}\right)^2} = \sqrt{1 + (x^2 + 1)} = \sqrt{x^2 + 2} \]

### Step 6: Find \( \sin(\phi) \)
Now, we can find \( \sin(\phi) \):
\[ \sin(\phi) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 2}} \]

### Step 7: Conclusion
Thus, the value of \( \sin\left(\cot^{-1}\left(\cos\left(\tan^{-1} x\right)\right)\right) \) is:
\[ \sin\left(\cot^{-1}\left(\cos\left(\tan^{-1} x\right)\right)\right) = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 2}} \]

### Final Answer
\[ \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 2}} \]

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