Question

Вычислите tgx+ctgx если x=15°

Answer 1

$bf tg(15^{circ})=tg left(dfrac{30^{circ}}{2} right) =dfrac{1-cos(30^{circ})}{sin(30^{circ})}=dfrac{1-frac{sqrt{3}}{2}}{frac{1}{2}}=2-sqrt{3} \ \ tgx+ctgx=dfrac{tg^2x+1}{tgx}=dfrac{(2-sqrt{3})^2+1}{2-sqrt{3}}=dfrac{8-4sqrt{3}}{2-sqrt{3}}=4$