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If α and β Are Acute Angles Satisfying Cos 2 α = 3 Cos 2 β − 1 3 − Cos 2 β , Then Tan α = - Mathematics

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Question

If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =

 

Options

  • \[\sqrt{2} \tan \beta\]

  • \[\frac{1}{\sqrt{2}}\tan \beta\]

  • \[\sqrt{2} \cot \beta\]

  • \[\frac{1}{\sqrt{2}} \cot \beta\]

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Solution

\[\sqrt{2} \tan \beta\]

\[\text{ Given } : \]

\[ \cos2\alpha = \frac{3\cos 2\beta - 1}{3 - \cos2\beta}\]

\[ \Rightarrow \frac{\cos2\alpha - 1}{\cos2\alpha + 1} = \frac{\left( 3\cos 2\beta - 1 \right) - \left( 3 - \cos2\beta \right)}{\left( 3\cos 2\beta - 1 \right) + \left( 3 - \cos2\beta \right)} \left( \text{ Using componendo and dividendo } \right)\]

\[ \Rightarrow \frac{\cos2\alpha - 1}{\cos2\alpha + 1} = \frac{4\cos 2\beta - 4}{2\cos 2\beta + 2}\]

\[ \Rightarrow - \frac{1 - \cos2\alpha}{1 + \cos2\alpha} = \frac{- 4\left( 1 - \cos 2\beta \right)}{2\left( 1 + \cos 2\beta \right)}\]

\[ \Rightarrow \frac{1 - \cos2\alpha}{1 + \cos2\alpha} = \frac{2\left( 1 - \cos 2\beta \right)}{\left( 1 + \cos 2\beta \right)}\]

\[ \Rightarrow \frac{2 \sin^2 \alpha}{2 \cos^2 \alpha} = \frac{2\left( 2 \sin^2 \beta \right)}{\left( 2 \cos^2 \beta \right)}\]

\[ \Rightarrow \tan^2 \alpha = 2 \tan^2 \beta\]

\[ \therefore \tan \alpha = \sqrt{2} \tan \beta\]

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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 44]

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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 25 | Page 44

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