Question

If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then

\[\cos2\alpha\]   is equal to

 

Options

• \[\sin2\beta\]

• \[\sin4\beta\]

   

• \[\sin3\beta\]

   

• \[\cos2\beta\]

   

Answer 1

It is given that \[\tan\alpha = \frac{1}{7}\]  and \[\tan\beta = \frac{1}{3}\]

Now,

\[\tan2\beta = \frac{2\tan\beta}{1 - \tan^2 \beta}\]

\[ = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}}\]

\[ = \frac{\frac{2}{3}}{\frac{8}{9}}\]

\[ = \frac{3}{4}\]

\[\therefore \tan\left( \alpha + 2\beta \right) = \frac{\tan\alpha + \tan2\beta}{1 - \tan\alpha \tan2\beta}\]
\[ = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7} \times \frac{3}{4}}\]
\[ = \frac{\frac{25}{28}}{\frac{25}{28}}\]
\[ = 1\]

\[\tan\left( \alpha + 2\beta \right) = 1 = \tan\frac{\pi}{4}\]

\[ \Rightarrow \alpha + 2\beta = \frac{\pi}{4}\]

\[ \Rightarrow \alpha = \frac{\pi}{4} - 2\beta\]

\[ \Rightarrow 2\alpha = \frac{\pi}{2} - 4\beta\]

\[ \Rightarrow \cos2\alpha = \cos\left( \frac{\pi}{2} - 4\beta \right) = \sin4\beta\]

\[\therefore \cos2\alpha = \sin4\beta\]