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

Vậy $\cos^2\alpha+\tan^2\alpha.\cos^2\alpha=1$ 

b/ $(\cos\alpha+\sin\alpha)^2-(\sin\alpha-\cos\alpha)^2+1-4\cos\alpha.\sin\alpha\\=(\cos\alpha+\sin\alpha-\sin\alpha+\cos\alpha)(\cos\alpha+\sin\alpha+\sin\alpha-\cos\alpha)+1-4\cos\alpha.\sin\alpha\\=2\cos\alpha.2\sin\alpha+1-4\cos\alpha.\sin\alpha\\=4\cos\alpha.\sin\alpha+1-4\cos\alpha.\sin\alpha\\=1$

Vậy biểu thức không phụ thuộc vào $\alpha$

Hướng dẫn trả lời:

a) `cos^2\alpha + tan^2\alpha.cos^2\alpha.`

`= cos^2\alpha + sin^2\alpha/cos^2\alpha.cos^2\alpha.`

`= cos^2\alpha + sin^2\alpha.`

`= 1.`

$\text{Vậy $cos^2\alpha + tan^2\alpha.cos^2\alpha$ = 1.}$

b) `(cos\alpha + sin\alpha)^2 - (sin\alpha + cos\alpha)^2 + 1 - 4cos\alpha.sin\alpha.`

`= (cos^2\alpha + 2sin\alphacos\alpha+ sin\alpha^2) - (sin^2\alpha + 2sin\alphacos\alpha+ cos^2\alpha) + 1 - 4sin\alphacos\alpha.`

`= cos^2\alpha + 2sin\alphacos\alpha+ sin\alpha^2 - sin^2\alpha - 2sin\alphacos\alpha - cos^2\alpha + 1 - 4sin\alphacos\alpha.`

`= (cos^2\alpha - cos^2\alpha) + (sin\alpha^2 - sin^2\alpha) + (2sin\alphacos\alpha - 2sin\alphacos\alpha - 4sin\alphacos\alpha) + 1.`

`= 1.`

$\text{Vậy biểu thức trên không phụ thuộc vào $\alpha$ (với $\alpha$ là góc nhọn tùy ý).}$