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)