a/ $\sin ^4 \alpha+\cos^4\alpha+2\sin^2\alpha\cos^2\alpha\\=(\sin^2\alpha+\cos^2\alpha)^2\\=1^2\\=1$
Vậy $\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cos^2\alpha=1$
b/ $(1-\cos\alpha)(1+\cos\alpha)\\=1^2-\cos^2\alpha\\=1-\cos^2\alpha\\=\sin^2\alpha+\cos^2\alpha-\cos^2\alpha\\=\sin^2\alpha$
Vậy $(1-\cos\alpha)(1+\cos\alpha)=\sin^2\alpha$
c/ $\cos^2\alpha+\tan^2\alpha.\cos^2\alpha\\=\cos^2\alpha+\left(\dfrac{\sin\alpha}{\cos\alpha}\right)^2.\cos^2\alpha\\=\cos^2\alpha+\dfrac{\sin^2\alpha}{\cos^2\alpha}.\cos^2\alpha\\=\cos^2\alpha+\sin^2\alpha\\=1$
Vậy $\cos^2\alpha+\tan^2\alpha.\cos^2\alpha=1$
