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

Options

A. x
B. `pi/2`
C. 1
D. none of these

Answer

To solve the problem \( \sin \left( \cot^{-1} \left[ \tan \left( \cos^{-1} x \right) \right] \right) \), we will break it down step by step.

### Step 1: Simplify \( \tan(\cos^{-1} x) \)

Let \( \theta = \cos^{-1} x \). Then, by definition of cosine, we have:
\[ \cos \theta = x \]
To find \( \tan \theta \), we need to find the opposite side of the triangle formed. Using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
we can find \( \sin \theta \):
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - x^2 \implies \sin \theta = \sqrt{1 - x^2} \]
Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{1 - x^2}}{x} \]

### Step 2: Substitute into \( \cot^{-1} \)

Now we substitute this into \( \cot^{-1} \):
\[ \cot^{-1} \left( \tan(\cos^{-1} x) \right) = \cot^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) \]
Using the identity \( \cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right) \):
\[ \cot^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right) \]

### Step 3: Find \( \sin \left( \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right) \right) \)

Let \( \phi = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right) \). We need to find \( \sin \phi \):
\[ \tan \phi = \frac{x}{\sqrt{1 - x^2}} \]
This implies that in a right triangle, the opposite side is \( x \) and the adjacent side is \( \sqrt{1 - x^2} \). The hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{x^2 + (1 - x^2)} = \sqrt{1} \]
Thus, \( h = 1 \).

Now we can find \( \sin \phi \):
\[ \sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{1} = x \]

### Final Answer

Therefore, we have:
\[ \sin \left( \cot^{-1} \left[ \tan \left( \cos^{-1} x \right) \right] \right) = x \]