Как решить, (sin 30' + cos 30')^2- tg 30' ( ' -градус)

(sin30°+cos30°)²-tg30°=sin²30°+2.sin30°.cos²30°+cos²30°-tg30°=
=(sin²30°+cos²30°)+(2.sin30°.cos30°)-tg30°=
=1+sin(2.30°)-tg30°=1+sin60°-tg30°=1+√3/2-√3/3=
=1+(3.√3-2.√3)/6=1+√3/6