Câu 1.$VT=\frac{cosx-sinx}{cosx+sinx}=\frac{\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx}{\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx}$$=\frac{sin\frac{\pi}{4}.cosx-cos\frac{\pi}{4}.sinx}{cos\frac{\pi}{4}.cosx+sin\frac{\pi}{4}.sinx}$$=\frac{sin(\frac{\pi}{4}-x)}{cos(\frac{\pi}{4}-x)}=tan(\frac{\pi}{4}-x)=VP$(đpcm)Câu 2.$C=\frac{-(2cos^22x-cos2x-1)}{sin2x+2sin2x.cos2x}=\frac{-(cos2x-1)(1+2cos2x)}{sin2x(1+2cos2x)}=\frac{1-cos2x}{sin2x}$$=\frac{2sin^2x}{2sinx.cosx}=\frac{sinx}{cosx}=tanx$
Câu 1.$VT=\frac{cosx-sinx}{cosx+sinx}=\frac{\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx}{\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx}$$=\frac{sin\frac{\pi}{4}.cosx-cos\frac{\pi}{4}.sinx}{cos\frac{\pi}{4}.cosx+sin\frac{\pi}{4}.sinx}$$=\frac{sin(\frac{\pi}{4}-x)}{cos(\frac{\pi}{4}-x)}=tan(\frac{\pi}{4}-x)=VP$(đpcm)Câu 2.$C=\frac{2cos^22x-cos2x-1}{sin2x+2sin2x.cos2x}=\frac{(cos2x-1)(1+2cos2x)}{sin2x(1+2cos2x)}=\frac{cos2x-1}{sin2x}$$=\frac{-2sin^2x}{2sinx.cosx}=\frac{-sinx}{cosx}=-tanx$
Câu 1.$VT=\frac{cosx-sinx}{cosx+sinx}=\frac{\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx}{\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx}$$=\frac{sin\frac{\pi}{4}.cosx-cos\frac{\pi}{4}.sinx}{cos\frac{\pi}{4}.cosx+sin\frac{\pi}{4}.sinx}$$=\frac{sin(\frac{\pi}{4}-x)}{cos(\frac{\pi}{4}-x)}=tan(\frac{\pi}{4}-x)=VP$(đpcm)Câu 2.$C=\frac{
-(2cos^22x-cos2x-1
)}{sin2x+2sin2x.cos2x}=\frac{
-(cos2x-1)(1+2cos2x)}{sin2x(1+2cos2x)}=\frac{
1-cos2x}{sin2x}$$=\frac{2sin^2x}{2sinx.cosx}=\frac{sinx}{cosx}=tanx$