Question

sin510° +√2 cos (-855°)+tg1125°+√3 ctg(-1380°)
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Answer 1

$sin510^\circ +\sqrt2\, cos(-855^\circ )+tg1125^\circ +\sqrt3\, ctg(-1380^\circ )=\\\\=sin(360^\circ +150^\circ )+\sqrt2\, cos(2\cdot 360^\circ +135^\circ )+tg(6\cdot 180^\circ +45^\circ )-\\\\-\sqrt3\, ctg(7\cdot 180^\circ +120^\circ )=\\\\=sin150^\circ +\sqrt2\, cos135^\circ +tg45^\circ -\sqrt3\, ctg120^\circ =\\\\=sin(180^\circ -30^\circ )+\sqrt2\, cos(90^\circ +45^\circ )+1-\sqrt3\, ctg(90^\circ +30^\circ )=\\\\=sin30^\circ -\sqrt2\, sin45^\circ +1+\sqrt3\, ctg30^\circ =$

$=\dfrac{1}{2}-\sqrt2\cdot \dfrac{\sqrt2}{2}+1+\sqrt3\cdot \sqrt3=\dfrac{1}{2}-1+1+3=3,5$