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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions  Exercise Multiple Choice Questions Question 14 Maths Textbook Solution.

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Answer: \frac{\pi}{6}

Hint: Take cos function to the RHS, so the variables get free.

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

We have to compute \alpha-\beta


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

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

                \begin{aligned} &=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ &=\frac{\pi}{6} \end{aligned}       

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