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

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