Question

If $\alpha ={\tan }^{-1}\left(\frac{\sqrt{3}x}{2y-x}\right),\beta ={\tan }^{-1}\left(\frac{2x-y}{\sqrt{3}y}\right)$, then find the value of: $\alpha -\beta$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $-\frac{\pi }{3}$

Answer 1

The correct option is A $\frac{\pi }{6}$

Given, $\alpha ={\tan }^{-1}\left(\frac{\sqrt{3}x}{2y-x}\right),\beta ={\tan }^{-1}\left(\frac{2x-y}{\sqrt{3}y}\right)$

$\alpha -\beta ={\tan }^{-1}\left(\frac{\sqrt{3}x}{2y-x}\right)-{\tan }^{-1}\left(\frac{2x-y}{\sqrt{3}y}\right)$

We know that

${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$

$={\tan }^{-1}\left(\frac{\frac{\sqrt{3}x}{2y-x}-\frac{2x-y}{\sqrt{3}y}}{1+\left(\frac{\sqrt{3}x}{2y-x}\right)\left(\frac{2x-y}{\sqrt{3}y}\right)}\right)$

$={\tan }^{-1}\left(\frac{\frac{3xy-4xy+2y^2+2x^2-xy}{\sqrt{3}y(2y-x)}}{\frac{2\sqrt{3}{y}^2-xy\sqrt{3}+2\sqrt{3}{x}^2-\sqrt{3}xy}{\sqrt{3}y(2y-x)}}\right)$

$={\tan }^{-1}\left(\frac{\frac{2y^2+2x^2-2xy}{\sqrt{3}y(2y-x)}}{\frac{2\sqrt{3}{y}^2+2\sqrt{3}{x}^2-2\sqrt{3}xy}{\sqrt{3}y(2y-x)}}\right)$

$={\tan }^{-1}\left(\frac{2(x^2+y^2-xy)}{2\sqrt{3}(y^2+x^2-xy)}\right)$

$={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$=\frac{\pi }{6}$