Question

If \[\tan A = \frac{1}{7}\]  and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B 

 

 

Answer 1

Given: 

\[\tan A = \frac{1}{7}\]  and \[\tan B = \frac{1}{3}\]

Using the identity  \[\tan2B = \frac{2\text{ tan } B}{1 - \tan^2 B}\] , we get

\[\tan2B = \frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}} = \frac{3}{4}\] 

Now, using the identities 

\[\cos2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} \text{ and }  \sin4B = \frac{2\tan2B}{1 + \tan^2 2B}\] , we get

\[\cos2A = \frac{1 - \left( \frac{1}{7} \right)^2}{1 + \left( \frac{1}{7} \right)^2} \text{ and }  \sin4B = \frac{2 \times \frac{3}{4}}{1 + \left( \frac{3}{4} \right)^2}\]
\[ \Rightarrow \cos2A = \frac{48}{50} \text{ and }  \sin4B = \frac{2 \times \frac{3}{4} \times 16}{25}\]
\[ \Rightarrow \cos2A = \frac{24}{25} \text{ and }  \sin4B = \frac{24}{25}\]

∴ cos 2A = sin 4B