Let $\cos{(\alpha + \beta) = \frac{4}{5}}$ and let $\sin{(\alpha - \beta) = \frac{5}{13}}$, where $0 \leq \alpha,\ \beta \leq \frac{\pi}{4}$. Then $ an{2\alpha}$ is equal to

Question

Let $\cos{(\alpha + \beta) = \frac{4}{5}}$ and let $\sin{(\alpha - \beta) = \frac{5}{13}}$, where $0 \leq \alpha,\ \beta \leq \frac{\pi}{4}$. Then $	an{2\alpha}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$\cos{(\alpha + \beta) = \frac{4}{5} \Rightarrow \alpha + \beta \in}$ 1st quadrant and
$\sin{(\alpha - \beta) = \frac{5}{13}}$
$\Rightarrow \ \alpha - \beta \in$ 1st quadrant
$\Rightarrow \ \ 2\alpha = (\alpha + \beta) + (\alpha - \beta)$
$	herefore\ \ 	an{2\alpha = \frac{	an{(\alpha + \beta) + 	an{(\alpha - \beta)}}}{1 - 	an{(\alpha + \beta)	an{(\alpha - \beta)}}}}$
$= \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4}.\frac{5}{12}} = \frac{56}{33}$

Q. Let $\cos{(\alpha + \beta) = \frac{4}{5}}$ and let $\sin{(\alpha - \beta) = \frac{5}{13}}$, where $0 \leq \alpha,\ \beta \leq \frac{\pi}{4}$. Then $\tan{2\alpha}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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