The value of `sin^(-1) {( sin. pi/3) x/sqrt((x^(2) + k^(2) - kx))} - cos^(_1) {( cos. pi/6) x/sqrt((x^(2) + k^(2) - kx))}" , where" (k/2 lt x lt 2k, k gt 0)` is

Options

A. `tan^(-1) ((2x^(2) + sk - k^(2))/(x^(2) - 2xk + k^(2)))`
B. `tan^(-1) ((x^(2) + 2xk - 2k^(2))/(x^(2) -2xk - k^(2)))`
C. ` tan^(-1) ((x^(2)+2xk - 2k^(2))/( 2x^(2) - 2xk+ 2k^(2)))`
D. None of the above

Answer

To solve the given problem, we need to simplify the expression step by step.

### Step 1: Write down the expression
We start with the expression:
\[ \sin^{-1} \left( \frac{\sin \frac{\pi}{3} \cdot x}{\sqrt{x^2 + k^2 - kx}} \right) - \cos^{-1} \left( \frac{\cos \frac{\pi}{6} \cdot x}{\sqrt{x^2 + k^2 - kx}} \right) \]

### Step 2: Substitute the values of sine and cosine
We know:
\[ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \]
Thus, we can rewrite the expression as:
\[ \sin^{-1} \left( \frac{\frac{\sqrt{3}}{2} x}{\sqrt{x^2 + k^2 - kx}} \right) - \cos^{-1} \left( \frac{\frac{\sqrt{3}}{2} x}{\sqrt{x^2 + k^2 - kx}} \right) \]

### Step 3: Use the identity for sine and cosine inverse
We can use the identity:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
This means:
\[ \sin^{-1}(a) - \cos^{-1}(a) = \sin^{-1}(a) - \left(\frac{\pi}{2} - \sin^{-1}(a)\right) = 2\sin^{-1}(a) - \frac{\pi}{2} \]
Let \( a = \frac{\sqrt{3}}{2} \cdot \frac{x}{\sqrt{x^2 + k^2 - kx}} \).

### Step 4: Substitute back into the expression
Thus, we can rewrite our expression as:
\[ 2 \sin^{-1} \left( \frac{\sqrt{3}}{2} \cdot \frac{x}{\sqrt{x^2 + k^2 - kx}} \right) - \frac{\pi}{2} \]

### Step 5: Further simplify using double angle identity
Using the identity:
\[ 2 \sin^{-1}(x) = \cos^{-1}(2x^2 - 1) \]
we can express:
\[ 2 \sin^{-1} \left( \frac{\sqrt{3}}{2} \cdot \frac{x}{\sqrt{x^2 + k^2 - kx}} \right) = \cos^{-1} \left( 2 \left( \frac{\sqrt{3}}{2} \cdot \frac{x}{\sqrt{x^2 + k^2 - kx}} \right)^2 - 1 \right) \]

### Step 6: Calculate the argument of cosine
Calculating the argument:
\[ = \cos^{-1} \left( \frac{3x^2}{x^2 + k^2 - kx} - 1 \right) = \cos^{-1} \left( \frac{3x^2 - (x^2 + k^2 - kx)}{x^2 + k^2 - kx} \right) = \cos^{-1} \left( \frac{2x^2 + kx - k^2}{x^2 + k^2 - kx} \right) \]

### Step 7: Final expression
Thus, the expression simplifies to:
\[ \frac{\pi}{2} - \cos^{-1} \left( \frac{2x^2 + kx - k^2}{x^2 + k^2 - kx} \right) \]
This means the original expression evaluates to:
\[ \sin^{-1} \left( \frac{2x^2 + kx - k^2}{x^2 + k^2 - kx} \right) \]

### Conclusion
The final value of the expression is:
\[ \sin^{-1} \left( \frac{2x^2 + kx - k^2}{x^2 + k^2 - kx} \right) \]