मराठी

If α = Tan − 1 ( √ 3 X 2 Y − X ) , β = Tan − 1 ( 2 X − Y √ 3 Y ) , Then α − β = (A) π 6 (B) π 3 (C) π 2 (D) − π 3 - Mathematics

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प्रश्न

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 =

पर्याय

  • `pi/6`

  • `pi/3`

  • `pi/2`

  • `-pi/3`

MCQ
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उत्तर

(a) `pi/6`

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

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पाठ 4: Inverse Trigonometric Functions - Exercise 4.16 [पृष्ठ १२०]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 14 | पृष्ठ १२०

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