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

Options

A. 0
B. `1//sqrt5`
C. `2//sqrt5`
D. `sqrt3//2`

Answer

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

### Step 1: Rewrite the Equation
We start with the given equation:
\[ \sin^{-1} x + \cot^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{2} \]

### Step 2: Use the Property of Inverse Trigonometric Functions
We know from trigonometric identities that:
\[ \tan^{-1} a + \cot^{-1} a = \frac{\pi}{2} \]
Thus, we can rewrite \( \cot^{-1} \left( \frac{1}{2} \right) \) as:
\[ \cot^{-1} \left( \frac{1}{2} \right) = \tan^{-1} \left( 2 \right) \]
This gives us:
\[ \sin^{-1} x + \tan^{-1} \left( 2 \right) = \frac{\pi}{2} \]

### Step 3: Isolate \( \sin^{-1} x \)
Now, we can isolate \( \sin^{-1} x \):
\[ \sin^{-1} x = \frac{\pi}{2} - \tan^{-1} \left( 2 \right) \]

### Step 4: Use the Sine Function
Taking the sine of both sides:
\[ x = \sin\left(\frac{\pi}{2} - \tan^{-1} \left( 2 \right)\right) \]
Using the co-function identity \( \sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta) \), we have:
\[ x = \cos\left(\tan^{-1} \left( 2 \right)\right) \]

### Step 5: Find \( \cos(\tan^{-1}(2)) \)
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 \( 1 \). By the Pythagorean theorem, the hypotenuse \( h \) is:
\[ h = \sqrt{1^2 + 2^2} = \sqrt{5} \]
Thus:
\[ \cos(\tan^{-1}(2)) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}} \]

### Step 6: Conclusion
Therefore, the value of \( x \) is:
\[ x = \frac{1}{\sqrt{5}} \]

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