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1-cos80-3-sin80-




Question Number 178182 by mathlove last updated on 13/Oct/22
(1/(cos80))−((√3)/(sin80))=?
$$\frac{\mathrm{1}}{{cos}\mathrm{80}}−\frac{\sqrt{\mathrm{3}}}{{sin}\mathrm{80}}=? \\ $$
Commented by cortano1 last updated on 13/Oct/22
 (1/(sin 10°)) −((√3)/(cos  10°)) = ((cos  10°−(√3) sin 10)/(sin 10° cos 10°))   = ((4((1/2)cos 10°−((√3)/2)sin 10°))/(sin 20°))   = ((4(sin 30° cos 10°−cos 30° sin 10°))/(sin 20°))  = 4
$$\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{10}°}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{cos}\:\:\mathrm{10}°}\:=\:\frac{\mathrm{cos}\:\:\mathrm{10}°−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{10}}{\mathrm{sin}\:\mathrm{10}°\:\mathrm{cos}\:\mathrm{10}°} \\ $$$$\:=\:\frac{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{10}°−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$\:=\:\frac{\mathrm{4}\left(\mathrm{sin}\:\mathrm{30}°\:\mathrm{cos}\:\mathrm{10}°−\mathrm{cos}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \\ $$$$=\:\mathrm{4} \\ $$
Commented by mathlove last updated on 13/Oct/22
thanks sir
$${thanks}\:{sir} \\ $$

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