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

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

Aicharya · Accepted Answer

Given that $ \begin{aligned} & \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} \ & 	ext { Now, } 	an 2 \alpha=	an [(\alpha+\beta)+(\alpha-\beta)] \ & =\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} \cdot \frac{5}{12}}=\frac{(9+5) 4}{48-15}=\frac{56}{33} \ & \end{aligned} $

If $\cos (\alpha+\beta)=\frac{4}{5}, \sin (\alpha-\beta)=\frac{5}{13}$ and $\alpha, \beta$ between 0 and…

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