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If 2 Tan α = 3 Tan β , Then Tan ( α − β ) = - Mathematics

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Question

If  \[2 \tan \alpha = 3 \tan \beta, \text{ then }  \tan \left( \alpha - \beta \right) =\]

 

Options

  • \[\frac{\sin 2 \beta}{5 - \cos 2 \beta}\]

  • \[\frac{\cos 2 \beta}{5 - \cos 2 \beta}\]

     

  • \[\frac{\sin 2 \beta}{5 + \cos 2 \beta}\]

  •  none of these

MCQ
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Solution

\[\frac{\sin 2 \beta}{5 - \cos 2 \beta}\]

\[\text{ Given } : \]

\[2\tan\alpha = 3\tan\beta\]

\[\text{ Now } , \]

\[\tan\left( \alpha - \beta \right) = \frac{tan\alpha - tan\beta}{1 + \text{ tan } \alpha \text{ tan } \beta}\]

\[ = \frac{\frac{3}{2}tan\beta - tan\beta}{1 + \left( \frac{3}{2}tan\beta \right)tan\beta}\]

\[ = \frac{3tan\beta - 2tan\beta}{2 + 3 \tan^2 \beta}\]

\[ = \frac{tan\beta}{2 + 3 \tan^2 \beta}\]

\[ = \frac{\frac{sin\beta}{cos\beta}}{2 + 3\frac{\sin^2 \beta}{\cos^2 \beta}}\]

\[ = \frac{sin\beta cos\beta}{2 \cos^2 \beta + 3 \sin^2 \beta}\]

\[ = \frac{sin\beta cos\beta}{2 + \sin^2 \beta}\]

\[ = \frac{2sin\beta cos\beta}{4 + 2 \sin^2 \beta}\]

\[ = \frac{\sin2\beta}{4 + 1 - \cos2\beta}\]

\[ \therefore \tan\left( \alpha - \beta \right) = \frac{\sin2\beta}{5 - \cos2\beta}\]

 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.5 [Page 43]

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RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 9 | Page 43

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