Giải thích các bước giải:
$b) VT=\dfrac{\sin^2x-\cos^2x}{1+2\sin x\cos }\\
=\dfrac{(\sin x-\cos x)(\sin x+\cos x)}{\sin^2x+\cos^2x+2\sin x\cos }\\
=\dfrac{(\sin x-\cos x)(\sin x+\cos x)}{(\sin x+\cos x)^2 }\\
=\dfrac{\sin x-\cos x}{\sin x+\cos x}\\
=\dfrac{\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}}{\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}}\\
=\dfrac{\tan x-1}{\tan x+1}=VP\Rightarrow ĐPCM\\ b) VT=3-4\sin^2x\\
=3(\sin^2x+\cos^2x)-4\sin^2x\\
=3\sin^2x+3\cos^2x-4\sin^2x\\
=3\cos^2x-\sin^2x\\
=3\cos^2x-(1-\cos^2x)\\
=3\cos^2x-1+\cos^2x\\
=4\cos^2x-1=VP\Rightarrow đpcm\\ b) VT=\dfrac{\sin^2\alpha+2\cos^2\alpha-1}{\cos^2\alpha}\\
=\dfrac{\sin^2\alpha+2\cos^2\alpha-\sin^2\alpha-\cos^2\alpha}{\cos^2\alpha}\\
=\dfrac{\cos^2\alpha}{\cos^2\alpha}\\
=1$
