The problem involves finding $\tan(2x)$ given the values of $\cos(x-y) = \frac{4}{5}$ and $\sin(x+y) = \frac{5}{13}$. Using the identities for cosine and sine, we can derive the values of $\sin(2x)$ and $\cos(2x)$, which are essential for calculating $\tan(2x)$. The final result for $\tan(2x)$ is determined to be $\frac{12}{7}$ after applying the double angle formulas and simplifying the expressions.
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lfdahl said:My solution:
\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)=\frac{3}{5}\rightarrow \tan(x-y)=\frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}=\frac{56}{33}\]
kaliprasad said:Above solution is correct but one solution is missing