If `sin^(-1)x-cos^(-1)x=(pi)/(6)`, then what is the value of x?

Options

A. A) `x=-(1)/(2)`
B. B) `x=1`
C. C) `x=(1)/(2)`
D. D) `x=(sqrt3)/(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
We know the identity:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Using this identity, we can express \( \sin^{-1} x \) in terms of \( \cos^{-1} x \):
\[ \sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x \]

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

### Step 3: Isolate \( \cos^{-1} x \)
Now, we can isolate \( \cos^{-1} x \) by moving \( \frac{\pi}{2} \) to the right side:
\[ -2\cos^{-1} x = \frac{\pi}{6} - \frac{\pi}{2} \]
To combine the fractions, we convert \( \frac{\pi}{2} \) to have a common denominator of 6:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
So, we have:
\[ -2\cos^{-1} x = \frac{\pi}{6} - \frac{3\pi}{6} = -\frac{2\pi}{6} \]
This simplifies to:
\[ -2\cos^{-1} x = -\frac{\pi}{3} \]

### Step 4: Divide by -2
Dividing both sides by -2 gives:
\[ \cos^{-1} x = \frac{\pi}{6} \]

### Step 5: Take the cosine of both sides
Now, we take the cosine of both sides:
\[ x = \cos\left(\frac{\pi}{6}\right) \]

### Step 6: Evaluate \( \cos\left(\frac{\pi}{6}\right) \)
We know that:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus, we find:
\[ x = \frac{\sqrt{3}}{2} \]

### Final Answer
The value of \( x \) is:
\[ \boxed{\frac{\sqrt{3}}{2}} \]