Question Number 20640 by oyshi last updated on 30/Aug/17
$$\mathrm{sin}\:\theta={K}\mathrm{cos}\:\left(\theta−\alpha\right) \\ $$$${so}\:{proof}\:{it},\mathrm{cot}\:\theta=\frac{\mathrm{1}−{K}\mathrm{sin}\:\alpha}{{K}\mathrm{cos}\:\alpha} \\ $$
Answered by mrW1 last updated on 30/Aug/17
$$\mathrm{sin}\:\theta=\mathrm{K}\left(\mathrm{cos}\:\theta\:\mathrm{cos}\:\alpha+\mathrm{sin}\:\theta\:\mathrm{sin}\:\alpha\right) \\ $$$$\left(\mathrm{1}−\mathrm{Ksin}\:\alpha\right)\mathrm{sin}\:\theta=\mathrm{Kcos}\:\alpha\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{cot}\:\theta=\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}=\frac{\mathrm{1}−\mathrm{Ksin}\:\alpha}{\mathrm{Kcos}\:\alpha} \\ $$