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Question

Prove that $\frac{\mathrm{tan}A+\mathrm{tan}B}{\mathrm{tan}A-\mathrm{tan}B}=\frac{\mathrm{sin}\left(A+B\right)}{\mathrm{sin}\left(A-B\right)}$ .

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Solution

$\mathrm{LHS}=\frac{\mathrm{tan}A+\mathrm{tan}B}{\mathrm{tan}A-\mathrm{tan}B}\phantom{\rule{0ex}{0ex}}=\frac{\frac{\mathrm{sin}A}{\mathrm{cos}A}+\frac{\mathrm{sin}B}{\mathrm{cos}B}}{\frac{\mathrm{sin}A}{\mathrm{cos}A}-\frac{\mathrm{sin}B}{\mathrm{cos}B}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B}{\mathrm{cos}A\mathrm{cos}B}}{\frac{\mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B}{\mathrm{cos}A\mathrm{cos}B}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B}{\mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}\left(A+B\right)}{\mathrm{sin}\left(A-B\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.$

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