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If alpha=tan^(-1)((sqrt(3)x)/(2y-x)) , b...

If `alpha=tan^(-1)((sqrt(3)x)/(2y-x))` , `beta=tan^(-1)((2x-y)/(sqrt(3)y))` , then `alpha-beta=` `pi/6` (b) `pi/3` (c) `pi/2` (d) `-pi/3`

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To solve the problem, we need to find the value of \( \alpha - \beta \) where: \[ \alpha = \tan^{-1}\left(\frac{\sqrt{3}x}{2y - x}\right) \] \[ \beta = \tan^{-1}\left(\frac{2x - y}{\sqrt{3}y}\right) \] ...
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