Question

\(\frac{1}{\cos^2\left(x\right)}-\tan^2x-\sin^2x\)

Answer 1

\(\frac{1}{cos^2x}-tan^2x-sin^2x=\frac{sin^2x+cos^2x}{cos^2x}-tan^2x-sin^2x=\frac{sin^2x}{cos^2x}+1-\frac{sin^2x}{cos^2x}-sin^2x=sin^2x+cos^2x-sin^2x=cos^2x\)