Question

If α = ${\tan }^{-1}\left(\frac{\sqrt{3}x}{2y-x}\right),\mathrm{\beta }={\tan }^{-1}\left(\frac{2x-y}{\sqrt{3}y}\right),$ then α − β =
(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{3}$

(c) $\frac{\mathrm{\pi }}{2}$

(d) $-\frac{\mathrm{\pi }}{3}$

Answer 1

(a) $\frac{\mathrm{\pi }}{6}$


We have
α = ${\tan }^{-1}\left(\frac{\sqrt{3}x}{2y-x}\right),\mathrm{\beta }={\tan }^{-1}\left(\frac{2x-y}{\sqrt{3}y}\right)$
$$\begin{gathered} \mathrm{Now},\alpha -\beta ={\tan }^{-1}\left(\frac{\sqrt{3}x}{2y-x}\right)-{\tan }^{-1}\frac{2x-y}{\sqrt{3}y} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{\sqrt{3}x}{2y-x}}-{\displaystyle \frac{2x-y}{\sqrt{3}y}}}{1+{\displaystyle \frac{\sqrt{3}x}{2y-x}}\times {\displaystyle \frac{2x-y}{\sqrt{3}y}}}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{3xy-4xy+2y^2+2x^2-xy}{\sqrt{3}y\left(2y-x\right)}}}{{\displaystyle \frac{\sqrt{3}y\left(2y-x\right)+\sqrt{3}x\left(2x-y\right)}{\sqrt{3}y\left(2y-x\right)}}}\right) \\ ={\tan }^{-1}\left(\frac{3xy-4xy+2y^2+2x^2-xy}{2\sqrt{3}{y}^2-\sqrt{3}xy+2\sqrt{3}{x}^2-\sqrt{3}xy}\right) \\ ={\tan }^{-1}\left(\frac{2y^2+2x^2-2xy}{2\sqrt{3}{y}^2+2\sqrt{3}{x}^2-2\sqrt{3}xy}\right) \\ ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\mathrm{\pi }}{6} \end{gathered}$$