Question Number 21140 by oyshi last updated on 14/Sep/17
$${if}\:\mathrm{tan}\:\beta=\frac{\mathrm{2sin}\:\alpha\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\alpha+\gamma\right)} \\ $$$${so}\:{proof}\:\mathrm{cot}\:\gamma+\mathrm{cot}\:\alpha=\mathrm{2cot}\:\beta \\ $$
Commented by $@ty@m last updated on 15/Sep/17
$${Step}\:\mathrm{1}.\:{Take}\:{reciprocal}\:{of}\:{both}\:{sides}. \\ $$$${Step}\:\mathrm{2}.\:{Multiply}\:{both}\:{sides}\:{by}\:\mathrm{2} \\ $$$${Step}\:\mathrm{3}.\:{Apply}\:{formula}\:{for}\:\mathrm{sin}\:\left(\alpha+\gamma\right) \\ $$$${Step}\:\mathrm{4}.\:{Use}\:{identity}\:\frac{{A}+{B}}{{C}}=\frac{{A}}{{C}}+\frac{{B}}{{C}} \\ $$
Answered by myintkhaing last updated on 15/Sep/17
$$\frac{\mathrm{1}}{\mathrm{tan}\:\beta}\:=\:\frac{\mathrm{sin}\:\left(\alpha+\gamma\right)}{\mathrm{2sin}\:\alpha\mathrm{sin}\:\gamma} \\ $$$$\mathrm{cot}\:\beta\:=\:\frac{\mathrm{sin}\:\alpha\mathrm{cos}\:\gamma+\mathrm{cos}\:\alpha\mathrm{sin}\:\gamma}{\mathrm{2sin}\:\alpha\mathrm{sin}\:\gamma} \\ $$$$\mathrm{2cot}\:\beta\:=\:\mathrm{cot}\:\gamma+\mathrm{cot}\:\alpha \\ $$