The equation `sin^-1x-cos^-1x=cos^-1(sqrt3/2)` has

Options

A. no solution
B. unique solution
C. infinite number of solution
D. none of these

Answer

To solve the equation \( \sin^{-1} x - \cos^{-1} x = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2} \). Therefore, we can rewrite the equation as:
\[ \sin^{-1} x - \left( \frac{\pi}{2} - \sin^{-1} x \right) = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \]
This simplifies to:
\[ 2 \sin^{-1} x - \frac{\pi}{2} = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \]

### Step 2: Find \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \)
The value of \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \) is known to be \( \frac{\pi}{6} \). Thus, we have:
\[ 2 \sin^{-1} x - \frac{\pi}{2} = \frac{\pi}{6} \]

### Step 3: Solve for \( \sin^{-1} x \)
Rearranging gives:
\[ 2 \sin^{-1} x = \frac{\pi}{2} + \frac{\pi}{6} \]
To combine the fractions, we find a common denominator:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \quad \text{and} \quad \frac{\pi}{6} = \frac{\pi}{6} \]
Thus:
\[ 2 \sin^{-1} x = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]

### Step 4: Divide by 2
Now, divide both sides by 2:
\[ \sin^{-1} x = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Step 5: Solve for \( x \)
Taking the sine of both sides gives:
\[ x = \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \]

### Step 6: Consider the range of \( x \)
Since \( \sin^{-1} x \) is defined for \( x \) in the range \([-1, 1]\), we check if \( x = \frac{\sqrt{3}}{2} \) is valid. It is valid as \( \frac{\sqrt{3}}{2} \) lies within this range.

### Step 7: Check for additional solutions
The original equation can also yield negative values, hence we consider:
\[ x = -\frac{\sqrt{3}}{2} \]
This is also valid as it lies within the range of \( \sin^{-1} \).

### Conclusion
Thus, the solutions to the equation \( \sin^{-1} x - \cos^{-1} x = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \) are:
\[ x = \frac{\sqrt{3}}{2} \quad \text{and} \quad x = -\frac{\sqrt{3}}{2} \]