Question

Prove that: $\frac{1}{(\sec A+\tan A)}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{(\sec A-\tan A)}$

Answer 1

$\frac{1}{(\sec A+\tan A)}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{(\sec A-\tan A)}\therefore \frac{1}{(\sec A+\tan A)}-\frac{1}{\cos A}-\frac{1}{\cos A}+\frac{1}{(\sec A-\tan A)}=0LHS=\frac{\sec A-\tan A+\sec A+\tan A}{\sec {}^{2}A-\tan {}^{2}A}-\frac{2}{\cos A}=2\sec A-\frac{2}{\cos A}=\frac{2}{\cos A}-\frac{2}{\cos A}=0=RHS$