\(\cot135^o=\cot(90^o+45^o)=-\tg45^o=-1\\\\\sin210^o=\sin(180^o+30^o)=-\sin30^o=-\dfrac{1}{2}\\\\\cos225^o=\cos(180^o+45^o)=-\cos45^o=-\dfrac{\sqrt2}{2}\)
\(\cot^2135^o-\sin210^o+5\cos^2225^o=(-1)^2-\left(-\dfrac{1}{2}\right)+5\cdot\left(-\dfrac{\sqrt2}{2}\right)^2\\\\=1+\dfrac{1}{2}+5\cdot\dfrac{1}{2}=1\dfrac{1}{2}+2\dfrac{1}{2}=4\)
Answer: 4.
