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# The value of $\frac{2\mathrm{sin}\mathrm{x}}{\mathrm{sin}3\mathrm{x}}+\frac{\mathrm{tan}\mathrm{x}}{\mathrm{tan}3\mathrm{x}}$ is ______ .

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Solution

## $\begin{array}{l}\frac{2\mathrm{sin}\mathrm{x}}{\mathrm{sin}3\mathrm{x}}+\frac{\mathrm{tan}\mathrm{x}}{\mathrm{tan}3\mathrm{x}}\\ =\frac{2\mathrm{sin}\mathrm{x}}{\mathrm{sin}\mathrm{x}\cdot \left(3-4{\mathrm{sin}}^{2}\mathrm{x}\right)}+\frac{\mathrm{tan}\mathrm{x}}{\frac{\left(3\mathrm{tan}\mathrm{x}-{\mathrm{tan}}^{3}\mathrm{x}\right)}{\left(1-3{\mathrm{tan}}^{2}\mathrm{x}\right)}}\end{array}$$\begin{array}{l}=\frac{2}{3-4{\mathrm{sin}}^{2}\mathrm{x}}+\frac{\left(1-3{\mathrm{tan}}^{2}\mathrm{x}\right)}{\left(3-{\mathrm{tan}}^{2}\mathrm{x}\right)}\\ =\frac{2}{3-4{\mathrm{sin}}^{2}\mathrm{x}}+\frac{1-\left(\frac{3{\mathrm{sin}}^{2}\mathrm{x}}{1-{\mathrm{sin}}^{2}\mathrm{x}}\right)}{3-\left(\frac{{\mathrm{sin}}^{2}\mathrm{x}}{1-{\mathrm{sin}}^{2}\mathrm{x}}\right)}\\ =\frac{2}{3-4{\mathrm{sin}}^{2}\mathrm{x}}+\frac{\left(1-4{\mathrm{sin}}^{2}\mathrm{x}\right)}{\left(3-4{\mathrm{sin}}^{2}\mathrm{x}\right)}\\ =\frac{\left(3-4{\mathrm{sin}}^{2}\mathrm{x}\right)}{\left(3-4{\mathrm{sin}}^{2}\mathrm{x}\right)}=1\end{array}$

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