STATEMENT-1 : The solution of `sin^-1 6x+sin^-1 6sqrt3x=pi/2` is , `x= +- 1/12.` and STATEMENT - 2 As, `sin^-1 x` is defined for `|x| <= 1.`

Options

A. Statement -1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement -5
B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -5
C. Statement-1 is True, Statement -2 is False
D. Statement -1 is False, Statement -2 is True

Answer

To solve the equation \( \sin^{-1}(6x) + \sin^{-1}(6\sqrt{3}x) = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Understand the Equation
The equation states that the sum of two inverse sine functions equals \( \frac{\pi}{2} \). We know that \( \sin^{-1}(a) + \sin^{-1}(b) = \frac{\pi}{2} \) if and only if \( a = \cos^{-1}(b) \) or \( b = \cos^{-1}(a) \).

### Step 2: Set Up the Relationship
From the equation, we can set:
\[ \sin^{-1}(6x) + \sin^{-1}(6\sqrt{3}x) = \frac{\pi}{2} \]
This implies:
\[ \sin^{-1}(6\sqrt{3}x) = \cos^{-1}(6x) \]

### Step 3: Use the Identity
Using the identity \( \cos^{-1}(a) = \frac{\pi}{2} - \sin^{-1}(a) \), we can rewrite:
\[ \sin^{-1}(6\sqrt{3}x) = \frac{\pi}{2} - \sin^{-1}(6x) \]
This means:
\[ 6\sqrt{3}x = \cos(\sin^{-1}(6x)) \]

### Step 4: Find the Cosine
Using the right triangle definition, we know:
\[ \cos(\sin^{-1}(y)) = \sqrt{1 - y^2} \]
Thus,
\[ \cos(\sin^{-1}(6x)) = \sqrt{1 - (6x)^2} = \sqrt{1 - 36x^2} \]

### Step 5: Set Up the Equation
Now we have:
\[ 6\sqrt{3}x = \sqrt{1 - 36x^2} \]

### Step 6: Square Both Sides
Squaring both sides gives:
\[ (6\sqrt{3}x)^2 = 1 - 36x^2 \]
\[ 108x^2 = 1 - 36x^2 \]

### Step 7: Rearrange the Equation
Combine like terms:
\[ 108x^2 + 36x^2 = 1 \]
\[ 144x^2 = 1 \]

### Step 8: Solve for x
Dividing both sides by 144:
\[ x^2 = \frac{1}{144} \]
Taking the square root:
\[ x = \pm \frac{1}{12} \]

### Conclusion
Thus, the solution to the equation \( \sin^{-1}(6x) + \sin^{-1}(6\sqrt{3}x) = \frac{\pi}{2} \) is:
\[ x = \pm \frac{1}{12} \]

### Verification of Statements
- **Statement 1** is true: The solution is indeed \( x = \pm \frac{1}{12} \).
- **Statement 2** is also true: The function \( \sin^{-1}(x) \) is defined for \( |x| \leq 1 \).