If `sin(cot^(-1)(x+1))=costan^(-1)x`, then x=

Options

A. `0`
B. `(9)/(4)`
C. `(1)/(2)`
D. `(-1)/(2)`

Answer

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

### Step 1: Understand the Inverse Functions
Let \( \theta = \cot^{-1}(x + 1) \). Then, we have:
\[ \cot(\theta) = x + 1 \]
This means that in a right triangle, the adjacent side is \( x + 1 \) and the opposite side is \( 1 \).

### Step 2: Find the Hypotenuse
Using the Pythagorean theorem, we can find the hypotenuse \( h \):
\[ h = \sqrt{(x + 1)^2 + 1^2} = \sqrt{(x + 1)^2 + 1} = \sqrt{x^2 + 2x + 2} \]

### Step 3: Calculate \( \sin(\theta) \)
Now, we can find \( \sin(\theta) \):
\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 2x + 2}} \]

### Step 4: Understand the Second Inverse Function
Let \( \alpha = \tan^{-1}(x) \). Then, we have:
\[ \tan(\alpha) = x \]
This means that in another right triangle, the opposite side is \( x \) and the adjacent side is \( 1 \).

### Step 5: Find the Hypotenuse for the Second Triangle
Using the Pythagorean theorem again, we can find the hypotenuse \( h' \):
\[ h' = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

### Step 6: Calculate \( \cos(\alpha) \)
Now, we can find \( \cos(\alpha) \):
\[ \cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 7: Set Up the Equation
Now we have:
\[ \sin(\cot^{-1}(x + 1)) = \cos(\tan^{-1}(x)) \]
This translates to:
\[ \frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 8: Cross Multiply
Cross-multiplying gives us:
\[ \sqrt{x^2 + 1} = \sqrt{x^2 + 2x + 2} \]

### Step 9: Square Both Sides
Squaring both sides results in:
\[ x^2 + 1 = x^2 + 2x + 2 \]

### Step 10: Simplify the Equation
Subtract \( x^2 \) from both sides:
\[ 1 = 2x + 2 \]
Now, subtract 2 from both sides:
\[ -1 = 2x \]
Finally, divide by 2:
\[ x = -\frac{1}{2} \]

### Final Answer
Thus, the solution to the equation is:
\[ \boxed{-\frac{1}{2}} \]