If `sin^(-1)x + cot^(-1)(1//2)=pi//2`, then what is the value of x ?

Options

A. 0
B. `1//sqrt(5)`
C. `2//sqrt(5)`
D. `sqrt(3)//2`

Explanation

To solve the equation \( \sin^{-1} x + \cot^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Understand the relationship between inverse trigonometric functions
We know that:
\[ \cot^{-1} y + \tan^{-1} y = \frac{\pi}{2} \]
This implies that:
\[ \cot^{-1} \left( \frac{1}{2} \right) = \tan^{-1} 2 \]
Thus, we can rewrite the original equation as:
\[ \sin^{-1} x + \tan^{-1} 2 = \frac{\pi}{2} \]

### Step 2: Rearranging the equation
From the above equation, we can isolate \( \sin^{-1} x \):
\[ \sin^{-1} x = \frac{\pi}{2} - \tan^{-1} 2 \]

### Step 3: Use the identity for sine
Using the identity \( \sin^{-1} a + \tan^{-1} b = \frac{\pi}{2} \) implies that:
\[ x = \cos(\tan^{-1} 2) \]

### Step 4: Calculate \( \cos(\tan^{-1} 2) \)
To find \( \cos(\tan^{-1} 2) \), we can use the right triangle definition:
- If \( \tan \theta = 2 \), then the opposite side is 2 and the adjacent side is 1.
- The hypotenuse can be calculated using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]
Thus, we have:
\[ \cos(\tan^{-1} 2) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{5}} \]

### Step 5: Conclusion
Therefore, the value of \( x \) is:
\[ x = \frac{1}{\sqrt{5}} \]

### Final Answer
\[ \boxed{\frac{1}{\sqrt{5}}} \]