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=$

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=$

Aicharya · Accepted Answer

$\begin{array}{ll}\cos (\alpha+\beta)=\frac{4}{5} & \Rightarrow 	an (\alpha+\beta)=\frac{3}{4} \ \sin (\alpha-\beta)=\frac{5}{13} & \Rightarrow 	an (\alpha-\beta)=\frac{5}{12} \ 	an 2 \alpha=	an (\alpha+\beta+\alpha-\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \frac{5}{12}}=\frac{56}{33}\end{array}$

Let $\cos (\alpha+\beta)=\frac{4}{5}$ and let $\sin (\alpha-\beta)=\frac{5}{13}$, where $0 \leq \alpha,…

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