If $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$, then $\alpha-\beta=$
Question.
If $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$, then $\alpha-\beta=$ (a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3}$
(C) $\frac{\pi}{2}$
(d) $-\frac{\pi}{3}$

Solution:
(a) $\frac{\pi}{6}$
We have
$\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$
Now, $\alpha-\beta=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right)-\tan ^{-1} \frac{2 x-y}{\sqrt{3} y}$
$=\tan ^{-1}\left(\frac{\frac{\sqrt{3} x}{2 y-x}-\frac{2 x-y}{\sqrt{3} y}}{1+\frac{\sqrt{3} x}{2 y-x} \times \frac{2 x-y}{\sqrt{3} y}}{2 y}\right)$
$=\tan ^{-1}\left(\frac{\frac{3 x y-4 x y+2 y^{2}+2 x^{2}-x y}{\sqrt{3} y(2 y-x)}}{\frac{\sqrt{3} y(2 y-x)+\sqrt{3} x(2 x-y)}{\sqrt{3} y(2 y-x)}}\right)$
$=\tan ^{-1}\left(\frac{3 x y-4 x y+2 y^{2}+2 x^{2}-x y}{2 \sqrt{3} y^{2}-\sqrt{3} x y+2 \sqrt{3} x^{2}-\sqrt{3} x y}\right)$
$=\tan ^{-1}\left(\frac{2 y^{2}+2 x^{2}-2 x y}{2 \sqrt{3} y^{2}+2 \sqrt{3} x^{2}-2 \sqrt{3} x y}\right)$
$=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$
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