Ответ: tgα = 3
cosα = 1/√10
sinα = 3/√10
Объяснение:
ctgα = 1/3
Так как угол острый, то значения всех выражений будут положительными.
1. tgα · ctgα = 1, ⇒ tgα = 1 / (ctgα) = 1 / (1/3) = 3
2. ![1+tg^{2}\alpha=\frac{1}{cos^{2}\alpha} 1+tg^{2}\alpha=\frac{1}{cos^{2}\alpha}](https://tex.z-dn.net/?f=1%2Btg%5E%7B2%7D%5Calpha%3D%5Cfrac%7B1%7D%7Bcos%5E%7B2%7D%5Calpha%7D)
![cos^{2}\alpha=\frac{1}{1+tg^{2}\alpha}=\frac{1}{1+9}=\frac{1}{10} cos^{2}\alpha=\frac{1}{1+tg^{2}\alpha}=\frac{1}{1+9}=\frac{1}{10}](https://tex.z-dn.net/?f=cos%5E%7B2%7D%5Calpha%3D%5Cfrac%7B1%7D%7B1%2Btg%5E%7B2%7D%5Calpha%7D%3D%5Cfrac%7B1%7D%7B1%2B9%7D%3D%5Cfrac%7B1%7D%7B10%7D)
![cos\alpha=\frac{1}{\sqrt{10}} cos\alpha=\frac{1}{\sqrt{10}}](https://tex.z-dn.net/?f=cos%5Calpha%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B10%7D%7D)
3. tgα = sinα / cosα, ⇒ sinα = tgα · cosα = 3 · 1/√10 = 3/√10