a, \(\dfrac{cosx}{1+sinx}+\dfrac{1+sinx}{cosx}=\dfrac{cos^2x+sin^2x+2sinx+1}{cosx\left(1+sinx\right)}\)
\(=\dfrac{2+2sinx}{cosx\left(1+sinx\right)}=\dfrac{2\left(1+sinx\right)}{cosx\left(1+sinx\right)}=\dfrac{2}{cosx}\)
Ta có đpcm
b, \(\dfrac{1}{cos^2x}-sin^2x-tan^2x=tan^2x+1-sin^2x-tan^2x=1-sin^2x=cos^2x\)
Vậy ta có đpcm