Answer :
[tex]Use:\\sin(x+y)=sinxcosy+sinycosx\\\\tan(-x)=-tanx\\tan(k\pi+x)=tanx\\\\cos(-x)=cosx\\cos(2k\pi+x)=cosx\\cos(x+y)=cosxcosy-sinxsiny[/tex]
[tex]sin\frac{3\pi}{2}=sin(\pi+\frac{\pi}{2})=sin\pi cos\frac{\pi}{2}+sin\frac{\pi}{2} cos\pi=0\cdot0+1\cdot(-1)=-1\\\\tan\left(-\frac{23\pi}{4}\right)=-tan\frac{23\pi}{4}=-tan\left[6\pi+\left(-\frac{1}{4}\pi\right)\right]=-tan\left(-\frac{1}{4}\pi\right)=tan\frac{\pi}{4}=1[/tex]
[tex]cos\left(-\frac{10\pi}{3}\right)=cos\frac{10\pi}{3}=cos\left(2\pi+1\frac{1}{3}\pi\right)=cos\left(1\frac{1}{3}\pi\right)=cos\left(\pi+\frac{\pi}{3}\right)\\=cos\pi cos\frac{\pi}{3}-sin\pi sin\frac{\pi}{3}=-1\cdot\frac{1}{2}-0\cdot\frac{\sqrt3}{2}=-\frac{1}{2}\\\\sin\frac{3\pi}{2}tan\left(-\frac{23\pi}{4}\right)-cos\left(-\frac{10\pi}{3}\right)=-1\cdot1-\left(-\frac{1}{2}\right)=-1+\frac{1}{2}=-\frac{1}{2}[/tex]
[tex]sin\frac{3\pi}{2}=sin(\pi+\frac{\pi}{2})=sin\pi cos\frac{\pi}{2}+sin\frac{\pi}{2} cos\pi=0\cdot0+1\cdot(-1)=-1\\\\tan\left(-\frac{23\pi}{4}\right)=-tan\frac{23\pi}{4}=-tan\left[6\pi+\left(-\frac{1}{4}\pi\right)\right]=-tan\left(-\frac{1}{4}\pi\right)=tan\frac{\pi}{4}=1[/tex]
[tex]cos\left(-\frac{10\pi}{3}\right)=cos\frac{10\pi}{3}=cos\left(2\pi+1\frac{1}{3}\pi\right)=cos\left(1\frac{1}{3}\pi\right)=cos\left(\pi+\frac{\pi}{3}\right)\\=cos\pi cos\frac{\pi}{3}-sin\pi sin\frac{\pi}{3}=-1\cdot\frac{1}{2}-0\cdot\frac{\sqrt3}{2}=-\frac{1}{2}\\\\sin\frac{3\pi}{2}tan\left(-\frac{23\pi}{4}\right)-cos\left(-\frac{10\pi}{3}\right)=-1\cdot1-\left(-\frac{1}{2}\right)=-1+\frac{1}{2}=-\frac{1}{2}[/tex]
