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Question Number 166718 by mathlove last updated on 26/Feb/22

csc10−(√3)sec10=?

$${csc}\mathrm{10}−\sqrt{\mathrm{3}}{sec}\mathrm{10}=? \\ $$

Commented by cortano1 last updated on 26/Feb/22

 (1/(sin 10°)) −((√3)/(cos 10°)) = ((cos 10°−(√3) sin 10°)/((1/2)sin 20°))   = ((2(2)((1/2)cos 10°−((√3)/2)sin 10°))/(sin 20°))   =((4(cos 60° cos 10°−sin 60° sin 10°))/(sin 20°))  = 4

$$\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\mathrm{10}°}\:=\:\frac{\mathrm{cos}\:\mathrm{10}°−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{10}°}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\:\frac{\mathrm{2}\left(\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{10}°−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\frac{\mathrm{4}\left(\mathrm{cos}\:\mathrm{60}°\:\mathrm{cos}\:\mathrm{10}°−\mathrm{sin}\:\mathrm{60}°\:\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$=\:\mathrm{4} \\ $$

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