If `0 lt x le 1, " then "`
` sqrt(1+x^2) [{x cos (cot^(-1)x)+sin(cot^(-1)x)}^(2)-1]^(1//2)=`

Options

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

Answer

To solve the given expression

\[ \sqrt{1+x^2} \left[ \left( x \cos(\cot^{-1} x) + \sin(\cot^{-1} x) \right)^2 - 1 \right]^{\frac{1}{2}}, \]

we will follow these steps:

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

### Step 2: Find the sine and cosine
From the definition of cotangent, we can express sine and cosine in terms of \(x\):
- \(\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{1}{x}\) implies that the opposite side is \(1\) and the adjacent side is \(x\).
- Using the Pythagorean theorem, the hypotenuse \(h\) can be calculated as:
\[ h = \sqrt{1^2 + x^2} = \sqrt{1 + x^2}. \]
- Therefore, we can find:
\[ \cos \theta = \frac{x}{\sqrt{1+x^2}}, \quad \sin \theta = \frac{1}{\sqrt{1+x^2}}. \]

### Step 3: Substitute sine and cosine into the expression
Now we substitute \(\cos(\cot^{-1} x)\) and \(\sin(\cot^{-1} x)\) into the expression:
\[ x \cos(\cot^{-1} x) + \sin(\cot^{-1} x) = x \cdot \frac{x}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+x^2}} = \frac{x^2 + 1}{\sqrt{1+x^2}}. \]

### Step 4: Square the expression
Now we need to square the expression:
\[ \left( x \cos(\cot^{-1} x) + \sin(\cot^{-1} x) \right)^2 = \left( \frac{x^2 + 1}{\sqrt{1+x^2}} \right)^2 = \frac{(x^2 + 1)^2}{1+x^2}. \]

### Step 5: Simplify the expression
Now we simplify:
\[ \left( x \cos(\cot^{-1} x) + \sin(\cot^{-1} x) \right)^2 - 1 = \frac{(x^2 + 1)^2 - (1+x^2)}{1+x^2} = \frac{x^4 + 2x^2 + 1 - 1 - x^2}{1+x^2} = \frac{x^4 + x^2}{1+x^2} = \frac{x^2(x^2 + 1)}{1+x^2}. \]

### Step 6: Take the square root
Now we take the square root:
\[ \sqrt{\left( x \cos(\cot^{-1} x) + \sin(\cot^{-1} x) \right)^2 - 1} = \sqrt{\frac{x^2(x^2 + 1)}{1+x^2}} = \frac{x\sqrt{x^2 + 1}}{\sqrt{1+x^2}} = x. \]

### Step 7: Multiply by \(\sqrt{1+x^2}\)
Finally, we multiply by \(\sqrt{1+x^2}\):
\[ \sqrt{1+x^2} \cdot x = x\sqrt{1+x^2}. \]

### Final Answer
Thus, the final result is:
\[ x \sqrt{1+x^2}. \]