Bài 1.5 trang 16

Đề bài

Chứng minh các đẳng thức:

a) ${\cos }^4\alpha -{\sin }^4\alpha =2{\cos }^2\alpha -1$;                

b) $\frac{{\cos }^2\alpha +{\tan }^2\alpha -1}{{\sin }^2\alpha }={\tan }^2\alpha$.

Lời giải chi tiết

a) Ta có: ${\cos }^4\alpha -{\sin }^4\alpha$

$=\left({\cos }^2\alpha -{\sin }^2\alpha \right)\left({\cos }^2\alpha +{\sin }^2\alpha \right)$

$={\cos }^2\alpha -{\sin }^2\alpha$

$={\cos }^2\alpha -(1-{\cos }^2\alpha )$

$={\cos }^2\alpha -1+{\cos }^2\alpha$

$=2{\cos }^2\alpha -1$ (đpcm).

b) Ta có: $\frac{{\cos }^2\alpha +{\tan }^2\alpha -1}{{\sin }^2\alpha }$

$=\frac{{\cos }^2\alpha +{\tan }^2\alpha -{\sin }^2\alpha -{\cos }^2\alpha }{{\sin }^2\alpha }$

$=\frac{{\tan }^2\alpha -{\sin }^2\alpha }{{\sin }^2\alpha }$

$=\frac{\frac{{\sin }^2\alpha }{{\cos }^2\alpha }-{\sin }^2\alpha }{{\sin }^2\alpha }$

$=\frac{1}{{\cos }^2\alpha }-1$

$={\tan }^2\alpha$ (đpcm).