Question

Вычислите tgx+ctgx если x=15°

Answer 1

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