Question-166686

Question Number 166686 by cortano1 last updated on 25/Feb/22

Commented by greogoury55 last updated on 26/Feb/22

(1/(1−sin^2 10°))+(2/(1−cos 40°)) +(2/(1−cos 80°))−2 = (1/(1−cos^2 80°))+(2/(1−cos 80°))+(2/(1−cos 40°))−2 = ((1+2(1+cos 80°))/(1−cos^2 80))+(2/(1−cos 40°))−2 = ((3+2cos 80°)/(1−(((1+cos 20°)/2))^2 ))+(2/(1−cos 40°))−2 =((12+8cos 80°)/(3−2cos 20°−cos^2 20°))+(2/(1−cos 40°))−2

$$\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\mathrm{10}°}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}\:+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{80}°}−\mathrm{2} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{80}°}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{80}°}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}−\mathrm{2} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{80}°\right)}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{80}}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}−\mathrm{2} \\ $$$$=\:\frac{\mathrm{3}+\mathrm{2cos}\:\mathrm{80}°}{\mathrm{1}−\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{20}°}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}−\mathrm{2} \\ $$$$=\frac{\mathrm{12}+\mathrm{8cos}\:\mathrm{80}°}{\mathrm{3}−\mathrm{2cos}\:\mathrm{20}°−\mathrm{cos}\:^{\mathrm{2}} \mathrm{20}°}+\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{40}°}−\mathrm{2} \\ $$$$ \\ $$

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