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Question Number 25771 by keyurpatel last updated on 14/Dec/17
show that ((cos 50)/(sin 40))+((sin 42)/(cos 48))−((2tan 18)/(cot 72))=0
$${show}\:{that}\:\frac{\mathrm{cos}\:\mathrm{50}}{\mathrm{sin}\:\mathrm{40}}+\frac{\mathrm{sin}\:\mathrm{42}}{\mathrm{cos}\:\mathrm{48}}−\frac{\mathrm{2tan}\:\mathrm{18}}{\mathrm{cot}\:\mathrm{72}}=\mathrm{0} \\ $$
Answered by deepak123 last updated on 14/Dec/17
sin (90−θ)=cosθ  cos (90−θ)=sin θ  tan (90−θ)=cot θ  apply  sin (90−50)/sin 40  +cos (90−42)/cos 48  −2cot (90−18)/cot 72=1+1−2=0
$$\mathrm{sin}\:\left(\mathrm{90}−\theta\right)=\mathrm{cos}\theta \\ $$$$\mathrm{cos}\:\left(\mathrm{90}−\theta\right)=\mathrm{sin}\:\theta \\ $$$$\mathrm{tan}\:\left(\mathrm{90}−\theta\right)=\mathrm{cot}\:\theta \\ $$$$\mathrm{apply} \\ $$$$\mathrm{sin}\:\left(\mathrm{90}−\mathrm{50}\right)/\mathrm{sin}\:\mathrm{40} \\ $$$$+\mathrm{cos}\:\left(\mathrm{90}−\mathrm{42}\right)/\mathrm{cos}\:\mathrm{48} \\ $$$$−\mathrm{2cot}\:\left(\mathrm{90}−\mathrm{18}\right)/\mathrm{cot}\:\mathrm{72}=\mathrm{1}+\mathrm{1}−\mathrm{2}=\mathrm{0} \\ $$

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