`sin{cot^(-1)[cos(tan^(-1)x)]}=....`

Options

A. `sqrt((x^(2)+2)/(x^(2)+1))`
B. `sqrt((x^(2)+1)/(x^(2)+2))`
C. `sqrt(x^(2)+1)`
D. none of these

Answer

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

### Step 1: Substitute \( \tan^{-1} x \) with \( \theta \)
Let \( \theta = \tan^{-1} x \). Then, we have:
\[ \tan \theta = x \]

### Step 2: Construct a right triangle
From the definition of tangent, we can construct a right triangle where the opposite side is \( x \) and the adjacent side is \( 1 \). Therefore, we can find the hypotenuse \( h \):
\[ h = \sqrt{x^2 + 1^2} = \sqrt{1 + x^2} \]

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

### Step 4: Substitute \( \cos \theta \) into the expression
Now we substitute \( \cos \theta \) into the original expression:
\[ \sin\left(\cot^{-1}\left[\cos\left(\tan^{-1} x\right)\right]\right) = \sin\left(\cot^{-1}\left[\frac{1}{\sqrt{1 + x^2}}\right]\right) \]

### Step 5: Let \( \phi = \cot^{-1}\left[\frac{1}{\sqrt{1 + x^2}}\right] \)
Now, let \( \phi = \cot^{-1}\left[\frac{1}{\sqrt{1 + x^2}}\right] \). Then:
\[ \cot \phi = \frac{1}{\sqrt{1 + x^2}} \]
Thus, \( \tan \phi = \sqrt{1 + x^2} \).

### Step 6: Construct another right triangle
We can construct another right triangle for \( \phi \) where the opposite side is \( \sqrt{1 + x^2} \) and the adjacent side is \( 1 \). The hypotenuse \( H \) is:
\[ H = \sqrt{(\sqrt{1 + x^2})^2 + 1^2} = \sqrt{1 + x^2 + 1} = \sqrt{2 + x^2} \]

### Step 7: Find \( \sin \phi \)
Now we can find \( \sin \phi \):
\[ \sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{1 + x^2}}{\sqrt{2 + x^2}} \]

### Step 8: Final expression
Thus, we have:
\[ \sin\left(\cot^{-1}\left[\cos\left(\tan^{-1} x\right)\right]\right) = \frac{\sqrt{1 + x^2}}{\sqrt{2 + x^2}} \]

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