If `sin (cot^(-1)(1-x))=cos(tan^(-1)(-x))`, then x is

Options

A. 1
B. `(1)/(2)`
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: Rewrite the left-hand side
Let \( \theta = \cot^{-1}(1-x) \). Then, we can express \( \sin(\theta) \) in terms of \( x \).

Using the identity:
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \]
In a right triangle where \( \cot(\theta) = 1-x \), we can set the adjacent side to \( 1-x \) and the opposite side to \( 1 \). The hypotenuse \( h \) can be calculated using the Pythagorean theorem:
\[ h = \sqrt{(1-x)^2 + 1^2} = \sqrt{(1-x)^2 + 1} \]
Thus, we have:
\[ \sin(\theta) = \frac{1}{\sqrt{(1-x)^2 + 1}} \]

### Step 2: Rewrite the right-hand side
Now, let \( \phi = \tan^{-1}(-x) \). Then, we can express \( \cos(\phi) \) in terms of \( x \).

Using the identity:
\[ \cos(\phi) = \frac{\text{adjacent}}{\text{hypotenuse}} \]
In a right triangle where \( \tan(\phi) = -x \), we can set the opposite side to \( -x \) and the adjacent side to \( 1 \). The hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{(-x)^2 + 1^2} = \sqrt{x^2 + 1} \]
Thus, we have:
\[ \cos(\phi) = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 3: Set the two expressions equal
Now we set the two expressions from Step 1 and Step 2 equal to each other:
\[ \frac{1}{\sqrt{(1-x)^2 + 1}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives:
\[ \sqrt{x^2 + 1} = \sqrt{(1-x)^2 + 1} \]
Squaring both sides results in:
\[ x^2 + 1 = (1-x)^2 + 1 \]
Expanding the right side:
\[ x^2 + 1 = 1 - 2x + x^2 + 1 \]
This simplifies to:
\[ x^2 + 1 = x^2 - 2x + 2 \]

### Step 5: Solve for \( x \)
Subtract \( x^2 \) from both sides:
\[ 1 = -2x + 2 \]
Rearranging gives:
\[ 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2} \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{\frac{1}{2}} \]