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प्रश्न
If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =
विकल्प
\[\sqrt{2} \tan \beta\]
\[\frac{1}{\sqrt{2}}\tan \beta\]
\[\sqrt{2} \cot \beta\]
\[\frac{1}{\sqrt{2}} \cot \beta\]
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उत्तर
\[\sqrt{2} \tan \beta\]
\[\text{ Given } : \]
\[ \cos2\alpha = \frac{3\cos 2\beta - 1}{3 - \cos2\beta}\]
\[ \Rightarrow \frac{\cos2\alpha - 1}{\cos2\alpha + 1} = \frac{\left( 3\cos 2\beta - 1 \right) - \left( 3 - \cos2\beta \right)}{\left( 3\cos 2\beta - 1 \right) + \left( 3 - \cos2\beta \right)} \left( \text{ Using componendo and dividendo } \right)\]
\[ \Rightarrow \frac{\cos2\alpha - 1}{\cos2\alpha + 1} = \frac{4\cos 2\beta - 4}{2\cos 2\beta + 2}\]
\[ \Rightarrow - \frac{1 - \cos2\alpha}{1 + \cos2\alpha} = \frac{- 4\left( 1 - \cos 2\beta \right)}{2\left( 1 + \cos 2\beta \right)}\]
\[ \Rightarrow \frac{1 - \cos2\alpha}{1 + \cos2\alpha} = \frac{2\left( 1 - \cos 2\beta \right)}{\left( 1 + \cos 2\beta \right)}\]
\[ \Rightarrow \frac{2 \sin^2 \alpha}{2 \cos^2 \alpha} = \frac{2\left( 2 \sin^2 \beta \right)}{\left( 2 \cos^2 \beta \right)}\]
\[ \Rightarrow \tan^2 \alpha = 2 \tan^2 \beta\]
\[ \therefore \tan \alpha = \sqrt{2} \tan \beta\]
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