Question

tg390°+ctg420°+tg480°+ctg510°=?

Answer 1

   
[tex]\displaystyle\\ \text{tg }390^o+\text{ctg }420^o+\text{tg }480^o+\text{ctg }510^o=\\\\ =\text{tg}(360^o+30^o)+\text{ctg}(360^o+60^o)+\\ +\text{tg}(360^o+120^o)+\text{ctg}(360+150^o)=\\\\ =\text{tg }30^o+\text{ctg }60^o+\text{tg }120^o+\text{ctg }150^o=\\\\ =\text{tg }30^o+\text{ctg }60^o+\text{tg}(180^o-60^o)+\text{ctg}(180^o -30^o)=\\\\ =\text{tg }30^o+\text{ctg }60^o-\text{tg }60^o-\text{ctg }30^o=\\\\ = \frac{ \sqrt{3} }{3} +\frac{ \sqrt{3} }{3}- \sqrt{3} -\sqrt{3}= [/tex]


[tex]\displaystyle\\ = \frac{ \sqrt{3} }{3} +\frac{ \sqrt{3} }{3}- \sqrt{3} -\sqrt{3}=\\\\ =2\left(\frac{ \sqrt{3} }{3}- \sqrt{3} \right)=2\left(\frac{ \sqrt{3} -3\sqrt{3}}{3}\right)=\\\\\\ =\frac{ 2\cdot (-2\sqrt{3})}{3}=\boxed{\bf\frac{-4\sqrt{3}}{3}}[/tex]