\(\frac{1}{cos^2x}-tan^2x=\frac{1}{cos^2x}-\frac{sin^2x}{cos^2x}=\frac{1-sin^2x}{cos^2x}=\frac{cos^2x}{cos^2x}=1\)
Bạn ghi đề ko đúng
Câu sau bạn cũng ghi đề ko đúng luôn, đề đúng phải là:
\(\frac{1-cosx}{sinx}\left(\frac{\left(1+cosx\right)^2}{sin^2x}-1\right)=2cotx\)
\(\frac{1-cosx}{sinx}\left(\frac{1+2cosx+cos^2x-sin^2x}{sin^2x}\right)=\frac{1-cosx}{sinx}\left(\frac{2cosx+2cos^2x}{sin^2x}\right)\)
\(=\frac{1-cosx}{sinx}\left[\frac{2cosx\left(cosx+1\right)}{sin^2x}\right]=\frac{\left(1-cos^2x\right).2cosx}{sinx.sin^2x}=\frac{sin^2x.2cosx}{sinx.sin^2x}=2cotx\)
\(\frac{tan^2x-sin^2x}{cot^2x-cos^2x}=\frac{sin^2x\left(\frac{1}{cos^2x}-1\right)}{cos^2x\left(\frac{1}{sin^2x}-1\right)}=\frac{sin^2x\left(\frac{1-cos^2x}{cos^2x}\right)}{cos^2x\left(\frac{1-sin^2x}{sin^2x}\right)}=\frac{sin^6x}{cos^6x}=tan^6x\)