If `A="tan"^(-1)((xsqrt3)/(2K-x))` and `B="tan"^(-1)((2x-K)/(Ksqrt3))`, then the value of `A-B` is

Options

A. `0^@`
B. `45^@`
C. `60^@`
D. `30^@`

Answer

To solve the problem, we need to find the value of \( A - B \) where:

\[ A = \tan^{-1}\left(\frac{x \sqrt{3}}{2K - x}\right) \]
\[ B = \tan^{-1}\left(\frac{2x - K}{K \sqrt{3}}\right) \]

We will use the identity for the difference of two inverse tangent functions:

\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right) \]

### Step 1: Apply the identity
We can apply the identity to find \( A - B \):

\[ A - B = \tan^{-1}\left(\frac{\frac{x \sqrt{3}}{2K - x} - \frac{2x - K}{K \sqrt{3}}}{1 + \left(\frac{x \sqrt{3}}{2K - x}\right)\left(\frac{2x - K}{K \sqrt{3}}\right)}\right) \]

### Step 2: Simplify the numerator
To simplify the numerator:

\[ \frac{x \sqrt{3}}{2K - x} - \frac{2x - K}{K \sqrt{3}} = \frac{x \sqrt{3} \cdot K \sqrt{3} - (2x - K)(2K - x)}{(2K - x)K \sqrt{3}} \]

Calculating the numerator:

\[ = \frac{3Kx - (4xK - 2x^2 + K^2)}{(2K - x)K \sqrt{3}} = \frac{3Kx - 4xK + 2x^2 - K^2}{(2K - x)K \sqrt{3}} = \frac{-Kx + 2x^2 - K^2}{(2K - x)K \sqrt{3}} \]

### Step 3: Simplify the denominator
Now, we simplify the denominator:

\[ 1 + \left(\frac{x \sqrt{3}}{2K - x}\right)\left(\frac{2x - K}{K \sqrt{3}}\right) = 1 + \frac{x(2x - K)}{(2K - x)K} \]

Combining terms gives:

\[ = \frac{(2K - x)K + x(2x - K)}{(2K - x)K} = \frac{2K^2 - Kx + 2x^2 - Kx}{(2K - x)K} = \frac{2K^2 + 2x^2 - 2Kx}{(2K - x)K} \]

### Step 4: Combine the results
Now we can combine the results from the numerator and denominator:

\[ A - B = \tan^{-1}\left(\frac{-Kx + 2x^2 - K^2}{2K^2 + 2x^2 - 2Kx}\right) \]

### Step 5: Factor out common terms
Factoring out common terms in the numerator and denominator:

\[ = \tan^{-1}\left(\frac{2x^2 - Kx - K^2}{2(x^2 - Kx + K^2)}\right) \]

### Step 6: Evaluate the expression
The expression simplifies to:

\[ = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Final Result
Thus, we find:

\[ A - B = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]