If $tan A = 1 / 2, tan B = 1 / 3,$ then show that $cos 2 A = sin 2 B$ .

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Solution

$tan A = \frac{1}{2}$
$cos 2 A = \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A} = \frac{1 - {(\frac{1}{2})}^{2}}{1 + {(\frac{1}{2})}^{2}} = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}$
For $tan B = \frac{1}{3}$
$sin 2 B = \frac{2 tan B}{1 + {tan}^{2} B} = \frac{2 \times \frac{1}{3}}{1 + {(\frac{1}{3})}^{2}} = \frac{\frac{2}{3}}{1 + \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{10}{9}} = \frac{6}{10} = \frac{3}{5}$
Hence $sin 2 B = cos 2 A$

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