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все что могу помочь! Думаю правильно на80%

[tex]\displaystyle\bf\\11.9\\\\Sin(270^\circ-\alpha )-Cos(180^\circ-\alpha )=-Cos\alpha -(-Cos\alpha)=\\\\=-Cos\alpha +Cos\alpha =0\\\\\\tg(90^\circ+\alpha )\cdot Ctg(270^\circ+\alpha )=-Ctg\alpha \cdot(-tg\alpha)=1\\\\\\tg\Big(\frac{\pi }{2} -\alpha \Big)-Ctg\Big(\pi -\alpha \Big)=Ctg\alpha -(-Ctg\alpha )=\\\\=Ctg\alpha +Ctg\alpha =2Ctg\alpha \\\\\\Sin(\pi +\alpha )\cdot tg\Big(\frac{3\pi }{2} +\alpha )=-Sin\alpha \cdot(-Ctg\alpha\Big)=[/tex]

[tex]\displaystyle\bf\\=-Sin\alpha \cdot\frac{Cos\alpha }{Sin\alpha } =Cos\alpha\\\\\\11.10\\\\tg(180^\circ+\alpha )-Ctg(270^\circ-\alpha )=-tg\alpha -tg\alpha =-2tg\alpha \\\\\\Sin(180^\circ-\alpha )\cdot Ctg(360^\circ+\alpha )=Sin\alpha \cdot Ctg\alpha =\\\\=Sin\alpha \cdot\frac{Cos\alpha }{Sin\alpha } =Cos\alpha \\\\\\Cos(2\pi -\alpha )-Sin\Big(\frac{3\pi }{2} -\alpha \Big)=Cos\alpha -(-Cos\alpha )=\\\\=Cos\alpha +Cos\alpha =2Cos\alpha[/tex]

[tex]\displaystyle\bf\\tg\Big(\frac{3\pi }{2} -\alpha \Big)\cdot Ctg\Big(\frac{\pi }{2} -\alpha \Big)=Ctg\alpha \cdot tg\alpha =1\\\\\\11.11\\\\Sin(\alpha- \pi )=-Sin(\pi- \alpha)=-Sin\alpha \\\\\\Cos\Big(\alpha -\frac{\pi }{2}\Big)=Cos\Big(\frac{\pi }{2}-\alpha \Big)=Sin\alpha \\\\\\tg(\alpha -2\pi )=-tg(2\pi -\alpha )=-(-tg\alpha)=tg\alpha \\\\\\Ctg\Big(\alpha -\frac{3\pi }{2}\Big)=-Ctg\Big(\frac{3\pi }{2} -\alpha \Big)=-tg\alpha[/tex]