Question
If
\[ \alpha = \tan^{-1}\left(\frac{\sqrt{3}x}{2y – x}\right), \quad \beta = \tan^{-1}\left(\frac{2x – y}{\sqrt{3}y}\right) \]
Find \( \alpha – \beta \).
Solution
Use identity:
\[ \tan(\alpha – \beta) = \frac{\tan\alpha – \tan\beta}{1 + \tan\alpha \tan\beta} \]
Substitute:
\[ \tan(\alpha – \beta) = \frac{\frac{\sqrt{3}x}{2y – x} – \frac{2x – y}{\sqrt{3}y}} {1 + \frac{\sqrt{3}x}{2y – x} \cdot \frac{2x – y}{\sqrt{3}y}} \]
Simplify numerator:
\[ = \frac{\frac{3xy – (2x – y)(2y – x)}{\sqrt{3}y(2y – x)}} {\frac{2y^2 – xy + 2x^2 – xy}{y(2y – x)}} \]
After simplification:
\[ \tan(\alpha – \beta) = \sqrt{3} \]
Thus,
\[ \alpha – \beta = \frac{\pi}{3} \]
(since principal value of tan⁻¹)
Final Answer:
\[ \boxed{\frac{\pi}{3}} \]
Key Concept
Apply tan(A − B) identity and simplify algebra carefully.