A value of x satisfying the equation `"sin"[cot^(-1)(1+x)]="cos"["tan"^(-1)x]` is:

Options

A. `-(1)/(2)`
B. -1
C. 0
D. `1/2`

Answer

To solve the equation \( \sin(\cot^{-1}(1+x)) = \cos(\tan^{-1}(x)) \), we will follow these steps:

### Step 1: Define the angles
Let:
- \( \alpha = \cot^{-1}(1+x) \)
- \( \beta = \tan^{-1}(x) \)

### Step 2: Construct triangles
For \( \alpha = \cot^{-1}(1+x) \):
- In a right triangle, if \( \cot(\alpha) = 1+x \), then the opposite side is 1 and the adjacent side is \( 1+x \).
- The hypotenuse can be calculated using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{1^2 + (1+x)^2} = \sqrt{1 + (1 + 2x + x^2)} = \sqrt{x^2 + 2x + 2} \]

For \( \beta = \tan^{-1}(x) \):
- In another right triangle, if \( \tan(\beta) = x \), then the opposite side is \( x \) and the adjacent side is 1.
- The hypotenuse can be calculated as:
\[ \text{Hypotenuse} = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

### Step 3: Find \( \sin(\alpha) \) and \( \cos(\beta) \)
Using the triangles constructed:
- For \( \sin(\alpha) \):
\[ \sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 2x + 2}} \]

- For \( \cos(\beta) \):
\[ \cos(\beta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 4: Set the two expressions equal
Now we set \( \sin(\alpha) \) equal to \( \cos(\beta) \):
\[ \frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ \sqrt{x^2 + 1} = \sqrt{x^2 + 2x + 2} \]

### Step 6: Square both sides
Squaring both sides results in:
\[ x^2 + 1 = x^2 + 2x + 2 \]

### Step 7: Simplify the equation
Subtract \( x^2 \) from both sides:
\[ 1 = 2x + 2 \]
Rearranging gives:
\[ 2x = 1 - 2 \implies 2x = -1 \implies x = -\frac{1}{2} \]

### Final Answer
The value of \( x \) satisfying the equation is:
\[ \boxed{-\frac{1}{2}} \]