If `sin^(-1)x + cot^(-1)(1/2)=(pi)/2` then `x` is equal to

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 will follow these steps:

### Step 1: Isolate \( \sin^{-1}x \)
We start by isolating \( \sin^{-1}x \) in the equation:
\[ \sin^{-1}x = \frac{\pi}{2} - \cot^{-1}\left(\frac{1}{2}\right) \]

### Step 2: Find \( \cot^{-1}\left(\frac{1}{2}\right) \)
Recall that \( \cot^{-1}(y) \) can be expressed in terms of tangent:
\[ \cot^{-1}\left(\frac{1}{2}\right) = \tan^{-1}(2) \]
This is because \( \cot(\theta) = \frac{1}{\tan(\theta)} \).

### Step 3: Substitute \( \tan^{-1}(2) \) into the equation
Now we substitute this back into our equation:
\[ \sin^{-1}x = \frac{\pi}{2} - \tan^{-1}(2) \]

### Step 4: Use the identity for \( \sin^{-1} \)
Using the identity \( \sin^{-1}y + \tan^{-1}y = \frac{\pi}{2} \), we can rewrite:
\[ \sin^{-1}x = \tan^{-1}(2) \]

### Step 5: Take the sine of both sides
Now we take the sine of both sides:
\[ x = \sin(\tan^{-1}(2)) \]

### Step 6: Use the definition of sine in terms of tangent
From the definition of sine and tangent:
\[ \sin(\tan^{-1}(y)) = \frac{y}{\sqrt{1+y^2}} \]
For \( y = 2 \):
\[ x = \frac{2}{\sqrt{1 + 2^2}} = \frac{2}{\sqrt{5}} \]

### Step 7: Final answer
Thus, the value of \( x \) is:
\[ x = \frac{2}{\sqrt{5}} \quad \text{or} \quad x = \frac{2\sqrt{5}}{5} \quad (\text{after rationalizing the denominator}) \]