If `A=tan^(-1) ""(xsqrt3)/(2lambda-x) and B=tan^(-1)""((2x-lambda)/(lambdasqrt3))` then the value of A-B is

Options

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

Answer

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

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

### Step 1: Use the formula for the difference of inverse tangents
We can use the formula for the difference of two inverse tangents:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \]
where \( x = \frac{x \sqrt{3}}{2\lambda - x} \) and \( y = \frac{2x - \lambda}{\lambda \sqrt{3}} \).

### Step 2: Calculate \( x - y \)
First, we find \( x - y \):
\[ x - y = \frac{x \sqrt{3}}{2\lambda - x} - \frac{2x - \lambda}{\lambda \sqrt{3}} \]
To combine these fractions, we need a common denominator:
\[ x - y = \frac{x \sqrt{3} \cdot \lambda \sqrt{3} - (2x - \lambda)(2\lambda - x)}{(2\lambda - x)(\lambda \sqrt{3})} \]

### Step 3: Expand the numerator
Now, we expand the numerator:
\[ x \sqrt{3} \cdot \lambda \sqrt{3} - (2x - \lambda)(2\lambda - x) \]
Calculating \( (2x - \lambda)(2\lambda - x) \):
\[ = 4\lambda x - 2x^2 - 2\lambda^2 + \lambda x \]
So, we have:
\[ x \sqrt{3} \cdot \lambda \sqrt{3} - (4\lambda x - 2x^2 - 2\lambda^2 + \lambda x) \]

### Step 4: Simplify the numerator
Combine like terms:
\[ = x \cdot 3\lambda - (4\lambda x - 2x^2 - 2\lambda^2 + \lambda x) = 3\lambda x - 4\lambda x + 2x^2 + 2\lambda^2 - \lambda x = -\lambda x + 2x^2 + 2\lambda^2 \]

### Step 5: Calculate \( 1 + xy \)
Next, we calculate \( 1 + xy \):
\[ xy = \left( \frac{x \sqrt{3}}{2\lambda - x} \right) \left( \frac{2x - \lambda}{\lambda \sqrt{3}} \right) = \frac{x(2x - \lambda)}{(2\lambda - x)(\lambda)} \]
Thus,
\[ 1 + xy = 1 + \frac{x(2x - \lambda)}{(2\lambda - x)(\lambda)} \]

### Step 6: Combine results
Now we can combine the results:
\[ A - B = \tan^{-1} \left( \frac{-\lambda x + 2x^2 + 2\lambda^2}{(2\lambda - x)(\lambda) + x(2x - \lambda)} \right) \]

### Step 7: Final simplification
After simplification, we find that:
\[ A - B = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \]
This corresponds to:
\[ A - B = \frac{\pi}{6} \]

### Conclusion
Thus, the value of \( A - B \) is:
\[ \frac{\pi}{6} \text{ or } 30^\circ \]