Question Number 20703 by tammi last updated on 01/Sep/17
$$\frac{\mathrm{sin}\alpha+\mathrm{sin}\:\mathrm{3}\alpha+\mathrm{sin5}\alpha}{\mathrm{cos}\:\alpha+\mathrm{cos}\:\mathrm{3}\alpha+\mathrm{cos}\:\mathrm{5}\alpha}=\mathrm{tan}\:\mathrm{3}\alpha \\ $$
Commented by myintkhaing last updated on 01/Sep/17
$$\mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{\mathrm{2sin3}\alpha\mathrm{cos2}\alpha+\mathrm{sin3}\alpha}{\mathrm{2cos3}\alpha\mathrm{cos2}\alpha+\mathrm{cos3}\alpha}=\frac{\mathrm{sin3}\alpha}{\mathrm{cos3}\alpha}=\mathrm{tan3}\alpha \\ $$$$\mathrm{proved} \\ $$
Answered by myintkhaing last updated on 01/Sep/17