Question Number 209781 by alusto22 last updated on 21/Jul/24
![((1+sin 40°−cos 40°)/(1+sin 40°+cos 40°)) =?](https://www.tinkutara.com/question/Q209781.png)
$$\:\:\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°−\mathrm{cos}\:\mathrm{40}°}{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°+\mathrm{cos}\:\mathrm{40}°}\:=? \\ $$
Answered by efronzo1 last updated on 21/Jul/24
![((1+sin 40°−(1−2sin^2 20°))/(1+sin 40°+2cos^2 20−1)) = ((2sin 20°(cos 20°+sin 20°))/(2cos 20°(cos 20°+sin 20°))) = tan 20°](https://www.tinkutara.com/question/Q209782.png)
$$\:\:\:\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{20}°\right)}{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°+\mathrm{2cos}\:^{\mathrm{2}} \mathrm{20}−\mathrm{1}}\: \\ $$$$\:\:=\:\frac{\mathrm{2sin}\:\mathrm{20}°\left(\mathrm{cos}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{20}°\right)}{\mathrm{2cos}\:\mathrm{20}°\left(\mathrm{cos}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{20}°\right)} \\ $$$$\:\:=\:\mathrm{tan}\:\mathrm{20}°\: \\ $$
Answered by Frix last updated on 22/Jul/24
![((1+sin 2x −cos 2x)/(1+sin 2x +cos 2x))=... =((2sin x cos x +2sin^2 x)/(2sin x cos x +2cos^2 x))= =((sin x)/(cos x))=tan x](https://www.tinkutara.com/question/Q209797.png)
$$\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}\:−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{2}{x}}=… \\ $$$$=\frac{\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:+\mathrm{2sin}^{\mathrm{2}} \:{x}}{\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:+\mathrm{2cos}^{\mathrm{2}} \:{x}}= \\ $$$$=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}=\mathrm{tan}\:{x} \\ $$