Question

Solve: $(\sec A+\cos A)(\sec A-\cos A)={\tan }^{2}A+{\sin }^{2}A$

Answer 1

$LHS=(\sec A+\cos A)(\sec A-\cos A)=\sec {}^{2}A-\sec A\cos A+\sec A\cos A-\cos {}^{2}A=\sec {}^{2}A-1+1-\cos {}^{2}A=\tan {}^{2}A+\sin {}^{2}A=RHS$