Câu 1.VT=cosx−sinxcosx+sinx=√22cosx−√22sinx√22cosx+√22sinx=sinπ4.cosx−cosπ4.sinxcosπ4.cosx+sinπ4.sinx=sin(π4−x)cos(π4−x)=tan(π4−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}$$=\frac{2sin^2x}{2sinx.cosx}=\frac{sinx}{cosx}=tanx$
Câu 1.VT=cosx−sinxcosx+sinx=√22cosx−√22sinx√22cosx+√22sinx=sinπ4.cosx−cosπ4.sinxcosπ4.cosx+sinπ4.sinx=sin(π4−x)cos(π4−x)=tan(π4−x)=VP(đpcm)Câu 2.$C=\frac{2cos^22x-cos2x-1}{sin2x+2sin2x.cos2x}=\frac{(cos2x-1)(1+2cos2x)}{sin2x(1+2cos2x)}=\frac{cos2x-1}{sin2x}$$=\frac{-2sin^2x}{2sinx.cosx}=\frac{-sinx}{cosx}=-tanx$
Câu 1.
VT=cosx−sinxcosx+sinx=√22cosx−√22sinx√22cosx+√22sinx=sinπ4.cosx−cosπ4.sinxcosπ4.cosx+sinπ4.sinx=sin(π4−x)cos(π4−x)=tan(π4−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$