Question

If α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right),\] 
 then α − β =

विकल्प

• `pi/6`

• `pi/3`

• `pi/2`

• `-pi/3`

Answer 1

(a) `pi/6`

We have
α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right)\]
\[\text{ Now }, \alpha - \beta = \tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right) - \tan^{- 1} \frac{2x - y}{\sqrt{3}y}\]
\[ = \tan^{- 1} \left( \frac{\frac{\sqrt{3}x}{2y - x} - \frac{2x - y}{\sqrt{3}y}}{1 + \frac{\sqrt{3}x}{2y - x} \times \frac{2x - y}{\sqrt{3}y}} \right)\]
\[ = \tan^{- 1} \left( \frac{\frac{3xy - 4xy + 2 y^2 + 2 x^2 - xy}{\sqrt{3}y\left( 2y - x \right)}}{\frac{\sqrt{3}y\left( 2y - x \right) + \sqrt{3}x\left( 2x - y \right)}{\sqrt{3}y\left( 2y - x \right)}} \right)\]
\[ = \tan^{- 1} \left( \frac{3xy - 4xy + 2 y^2 + 2 x^2 - xy}{2\sqrt{3} y^2 - \sqrt{3}xy + 2\sqrt{3} x^2 - \sqrt{3}xy} \right)\]
\[ = \tan^{- 1} \left( \frac{2 y^2 + 2 x^2 - 2xy}{2\sqrt{3} y^2 + 2\sqrt{3} x^2 - 2\sqrt{3}xy} \right)\]
\[ = \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}\]