If `sin^(-1)(tan(pi)/4)-sin^(-1)(sqrt(3/y))-(pi)/6=0` and `x^(2)=y` then `x` is equal to

Options

A. `2`
B. `4`
C. `sqrt(2)`
D. `sqrt(3)`

Explanation

To solve the equation \( \sin^{-1}(\tan(\pi/4)) - \sin^{-1}(\sqrt{3/y}) - \frac{\pi}{6} = 0 \) and find the value of \( x \) given that \( x^2 = y \), we can follow these steps:

### Step 1: Simplify \( \tan(\pi/4) \)
We know that:
\[ \tan(\pi/4) = 1 \]
Thus, we can rewrite the equation as:
\[ \sin^{-1}(1) - \sin^{-1}(\sqrt{3/y}) - \frac{\pi}{6} = 0 \]

### Step 2: Evaluate \( \sin^{-1}(1) \)
The value of \( \sin^{-1}(1) \) is:
\[ \sin^{-1}(1) = \frac{\pi}{2} \]
Now, substituting this back into the equation gives:
\[ \frac{\pi}{2} - \sin^{-1}(\sqrt{3/y}) - \frac{\pi}{6} = 0 \]

### Step 3: Rearranging the Equation
Now, rearranging the equation:
\[ \frac{\pi}{2} - \frac{\pi}{6} = \sin^{-1}(\sqrt{3/y}) \]
To simplify the left side, we need a common denominator. The least common multiple of 2 and 6 is 6:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
Thus,
\[ \frac{3\pi}{6} - \frac{\pi}{6} = \sin^{-1}(\sqrt{3/y}) \]
This simplifies to:
\[ \frac{2\pi}{6} = \sin^{-1}(\sqrt{3/y}) \]
or
\[ \frac{\pi}{3} = \sin^{-1}(\sqrt{3/y}) \]

### Step 4: Apply the Sine Function
Taking the sine of both sides:
\[ \sin\left(\frac{\pi}{3}\right) = \sqrt{3/y} \]
We know that:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus, we have:
\[ \frac{\sqrt{3}}{2} = \sqrt{3/y} \]

### Step 5: Square Both Sides
Squaring both sides gives:
\[ \left(\frac{\sqrt{3}}{2}\right)^2 = \left(\sqrt{3/y}\right)^2 \]
This simplifies to:
\[ \frac{3}{4} = \frac{3}{y} \]

### Step 6: Solve for \( y \)
Cross-multiplying gives:
\[ 3y = 12 \quad \Rightarrow \quad y = 4 \]

### Step 7: Find \( x \)
Given that \( x^2 = y \):
\[ x^2 = 4 \]
Taking the square root of both sides:
\[ x = \pm 2 \]

### Final Answer
Thus, the value of \( x \) is:
\[ x = 2 \quad \text{or} \quad x = -2 \]