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If $\cos(\alpha+\beta )=\frac{3}{5},\sin(\alpha-\beta )=\frac{5}{13}$ and $0<\alpha,\beta <\frac{\pi}{4}$ then $\tan(2\alpha)$ is equal to :

Option 1)
$\frac{21}{16}$

Option 2)
$\frac{63}{16}$

Option 3)
$\frac{33}{52}$

Option 4)
$\frac{63}{52}$

Answers (1)

best_answer

$\\\cos (\alpha +\beta )=\frac{3}{5}\\\; \; \; \\\sin (\alpha -\beta )=\frac{5}{13}$

$0<\alpha , \beta <\frac{4}{4}\; \; \; \; \; \; then\; \; \; \tan(2\alpha )=?$

$\\\cos(\alpha +\beta )=\frac{3}{5}\\\; \; \; \; \\\tan(\alpha +\beta )=\frac{4}{3}$

$\\\sin(\alpha -\beta )=\frac{5}{13}\\\; \; \; \; \\\tan(\alpha -\beta )=\frac{5}{12}$

$\\\alpha+\beta=\tan^{-1}\left ( \frac{4}{3} \right )\\\; \; \; \\\alpha-\beta=\tan^{-1}\left ( \frac{5}{12} \right )$

$=2\alpha=\tan^{-1}\left ( \frac{4}{3} \right )+\tan^{-1}\left ( \frac{5}{12} \right )$

$=\tan^{-1}\left ( \frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times\frac{5}{12}} \right )$

$=\tan^{-1}\left ( \frac{\frac{21}{12}}{\frac{36-20}{36}} \right )=\frac{\frac{21}{12}}{\frac{16}{36}}$

$=\tan^{-1}\left ( \frac{3\times 21}{16} \right )$

$=\tan^{-1}\left (\frac{63}{16} \right )$

Option 1)

$\frac{21}{16}$

Option 2)

$\frac{63}{16}$

Option 3)

$\frac{33}{52}$

Option 4)

$\frac{63}{52}$

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