Cho cota=1.Tìm sina,cosa,tana Giúp mình với!

Ta có: $\cot a = \frac{1}{\tan a}$, nên nếu $\cot a = 1$ thì $\tan a = 1$. Từ $\tan a = 1$, ta có: $\sin a = \cos a$. Áp dụng định lý Pytago: $\sin^2 a + \cos^2 a = 1$, ta có: $\sin^2 a + \sin^2 a = 1 \Rightarrow 2\sin^2 a = 1 \Rightarrow \sin^2 a = \frac{1}{2} \Rightarrow \sin a = \cos a = \frac{\sqrt{2}}{2}$. Vậy, $\sin a = \cos a = \frac{\sqrt{2}}{2}$, $\tan a = 1$. Tính giá trị của biểu thức: $K=\sin^272^0+\cos^225^0-\tan57^0+\sin^234^0+\tan63^0-12.\tan37^0.\tan53^0.$ $K=\left(\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2-\tan57^0+\left(\frac{\sqrt{2}}{2}\right)^2+\tan63^0-12.\tan37^0.\tan53^0.$ $K=\frac{1}{2}+\frac{1}{2}-\tan57^0+\frac{1}{2}+\tan63^0-12.\tan37^0.\tan53^0.$ $K=1-\tan57^0+\tan63^0-12.\tan37^0.\tan53^0.$ $K=1-\tan(90^0-33^0)+\tan(90^0-27^0)-12.\tan37^0.\tan53^0.$ $K=1-\cot33^0+\cot27^0-12.\tan37^0.\tan53^0.$ $K=1-\frac{1}{\tan33^0}+\frac{1}{\tan27^0}-12.\tan37^0.\tan53^0.$ $K=1-\frac{1}{\tan(90^0-57^0)}+\frac{1}{\tan(90^0-63^0)}-12.\tan37^0.\tan53^0.$ $K=1-\frac{1}{\cot57^0}+\frac{1}{\cot63^0}-12.\tan37^0.\tan53^0.$ $K=1-\frac{1}{\frac{1}{\tan57^0}}+\frac{1}{\frac{1}{\tan63^0}}-12.\tan37^0.\tan53^0.$ $K=1-\tan57^0+\tan63^0-12.\tan37^0.\tan53^0.$ $K=0$. $M=\frac{\sin27^0}{\cos27^0}-7\sin^265^0-7\sin^225^0-\tan27^0+\tan45^0$ $M=\tan27^0-7\left(\frac{\sqrt{2}}{2}\right)^2-7\left(\frac{\sqrt{2}}{2}\right)^2-\tan27^0+\tan45^0$ $M=\tan27^0-\frac{7}{2}- \frac{7}{2}-\tan27^0+1$ $M=-3+\frac{1}{\tan27^0}-\tan27^0$ $M=-3+\frac{1}{\tan(90^0-63^0)}-\tan(90^0-63^0)$ $M=-3+\frac{1}{\cot63^0}-\cot63^0$ $M=-3+\frac{1}{\frac{1}{\tan63^0}}-\frac{1}{\tan63^0}$ $M=-3+\tan63^0-\tan63^0$ $M=-3$. $N=12.\tan32^0.\tan58^0+\sin^251^0+\sin^239^0-\frac{8.\cot35^0}{\tan55^0}$. $N=12.\tan32^0.\tan(90^0-32^0)+\left(\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2-\frac{8.\frac{1}{\tan35^0}}{\tan(90^0-35^0)}$. $N=12.\tan32^0.\cot32^0+\frac{1}{2}+\frac{1}{2}-\frac{8.\frac{1}{\tan35^0}}{\cot35^0}$. $N=12-8=4$. Vậy, $K=0$, $M=-3$, $N=4$.