\text { If } \cos (\alpha+\beta)=4 / 5, \sin (\alpha-\beta)=5 / 13 \text { and } \alpha, \beta \text { lie between } 0 \text { and } \frac{\pi}{4}, \text { find } \tan 2 \alpha

Answers (1)

\tan 2 \alpha=\tan (\alpha+\beta+\alpha-\beta)=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)}

                                                                =\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4} \times \frac{12}{5}}=\frac{\frac{15+48}{20}}{\frac{20-36}{20}}

                                                                \tan 2 \alpha=\frac{63}{-16}

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