If `sin^(-1)x-cos^(-1)x=(pi)/(6), " then :" x=`

Options

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

Answer

To solve the equation \( \sin^{-1} x - \cos^{-1} x = \frac{\pi}{6} \), we can follow these steps:

### Step 1: Use the identity for sine and cosine inverse
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
From this identity, we can express \( \cos^{-1} x \) in terms of \( \sin^{-1} x \):
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]

### Step 2: Substitute into the equation
Now, substitute \( \cos^{-1} x \) into the original equation:
\[ \sin^{-1} x - \left( \frac{\pi}{2} - \sin^{-1} x \right) = \frac{\pi}{6} \]
This simplifies to:
\[ \sin^{-1} x - \frac{\pi}{2} + \sin^{-1} x = \frac{\pi}{6} \]
Combining like terms gives:
\[ 2 \sin^{-1} x - \frac{\pi}{2} = \frac{\pi}{6} \]

### Step 3: Isolate \( \sin^{-1} x \)
Add \( \frac{\pi}{2} \) to both sides:
\[ 2 \sin^{-1} x = \frac{\pi}{6} + \frac{\pi}{2} \]
To add these fractions, convert \( \frac{\pi}{2} \) to sixths:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
So,
\[ 2 \sin^{-1} x = \frac{\pi}{6} + \frac{3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]

### Step 4: Divide by 2
Now, divide both sides by 2:
\[ \sin^{-1} x = \frac{2\pi}{3} \cdot \frac{1}{2} = \frac{\pi}{3} \]

### Step 5: Find \( x \)
Now, take the sine of both sides:
\[ x = \sin\left(\frac{\pi}{3}\right) \]
We know that:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus, the value of \( x \) is:
\[ x = \frac{\sqrt{3}}{2} \]

### Final Answer
Therefore, the solution is:
\[ \boxed{\frac{\sqrt{3}}{2}} \]