Chứng minh đẳng thức lượng giác

a) Biến đổi vế trái: \[\cos^4\alpha - \sin^4\alpha = \left(\cos^2\alpha - \sin^2\alpha\right)\left(\cos^2\alpha + \sin^2\alpha\right)\] Vì \(\sin^2\alpha + \cos^2\alpha = 1\) nên: \[= \cos^2\alpha - \sin^2\alpha\] Thay \(\sin^2\alpha = 1 - \cos^2\alpha\): \[= \cos^2\alpha - (1 - \cos^2\alpha) = \cos^2\alpha - 1 + \cos^2\alpha = 2\cos^2\alpha - 1\] Vậy \(\cos^4\alpha - \sin^4\alpha = 2\cos^2\alpha - 1\) (đpcm). b) Biến đổi vế trái. Ở tử số, thay \(1 = \sin^2\alpha + \cos^2\alpha\): \[\frac{\cos^2\alpha + \tan^2\alpha - 1}{\sin^2\alpha} = \frac{\cos^2\alpha + \tan^2\alpha - \sin^2\alpha - \cos^2\alpha}{\sin^2\alpha} = \frac{\tan^2\alpha - \sin^2\alpha}{\sin^2\alpha}\] Thay \(\tan^2\alpha = \dfrac{\sin^2\alpha}{\cos^2\alpha}\): \[= \frac{\dfrac{\sin^2\alpha}{\cos^2\alpha} - \sin^2\alpha}{\sin^2\alpha} = \frac{1}{\cos^2\alpha} - 1\] Vì \(\dfrac{1}{\cos^2\alpha} - 1 = \dfrac{1 - \cos^2\alpha}{\cos^2\alpha} = \dfrac{\sin^2\alpha}{\cos^2\alpha} = \tan^2\alpha\), ta có: \[\frac{\cos^2\alpha + \tan^2\alpha - 1}{\sin^2\alpha} = \tan^2\alpha \text{ (đpcm).}\]