Find the value of, `cos [tan^(-1) {sin (cot^(-1)x)}]`

Answer

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

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

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

### Step 3: Find the hypotenuse
Using the Pythagorean theorem, we can find the hypotenuse \( h \):
\[ h = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

### Step 4: Find \( \sin \theta \)
Now, we can find \( \sin \theta \):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 5: Substitute into the cosine function
Now we need to evaluate \( \cos \left( \tan^{-1} \left( \sin \theta \right) \right) \). Let \( \alpha = \tan^{-1} \left( \sin \theta \right) \). Thus, we have:
\[ \sin \theta = \frac{1}{\sqrt{x^2 + 1}} \]
This means:
\[ \tan \alpha = \sin \theta = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 6: Construct another right triangle for \( \alpha \)
For the angle \( \alpha \):
- The opposite side is \( 1 \)
- The adjacent side is \( \sqrt{x^2 + 1} \)

### Step 7: Find the hypotenuse for \( \alpha \)
Using the Pythagorean theorem again:
\[ \text{Hypotenuse} = \sqrt{1^2 + \left( \sqrt{x^2 + 1} \right)^2} = \sqrt{1 + (x^2 + 1)} = \sqrt{x^2 + 2} \]

### Step 8: Find \( \cos \alpha \)
Now we can find \( \cos \alpha \):
\[ \cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 2}} \]

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