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Question
If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then
\[\cos2\alpha\] is equal to
Options
\[\sin2\beta\]
- \[\sin4\beta\]
- \[\sin3\beta\]
- \[\cos2\beta\]
Solution
It is given that \[\tan\alpha = \frac{1}{7}\] and \[\tan\beta = \frac{1}{3}\]
Now,
\[\tan2\beta = \frac{2\tan\beta}{1 - \tan^2 \beta}\]
\[ = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}}\]
\[ = \frac{\frac{2}{3}}{\frac{8}{9}}\]
\[ = \frac{3}{4}\]
\[\therefore \tan\left( \alpha + 2\beta \right) = \frac{\tan\alpha + \tan2\beta}{1 - \tan\alpha \tan2\beta}\]
\[ = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} \times \frac{3}{4}}\]
\[ = \frac{\frac{25}{28}}{\frac{25}{28}}\]
\[ = 1\]
\[\tan\left( \alpha + 2\beta \right) = 1 = \tan\frac{\pi}{4}\]
\[ \Rightarrow \alpha + 2\beta = \frac{\pi}{4}\]
\[ \Rightarrow \alpha = \frac{\pi}{4} - 2\beta\]
\[ \Rightarrow 2\alpha = \frac{\pi}{2} - 4\beta\]
\[ \Rightarrow \cos2\alpha = \cos\left( \frac{\pi}{2} - 4\beta \right) = \sin4\beta\]
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