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

$\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A+\tan A}$

$LHS\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}$

$\frac{1}{\sec A-\tan A}.\frac{\sec A+\tan A}{\sec A+\tan A}-\frac{1}{\cos A}$

$\frac{\sec A+\tan A}{{\sec }^{2}A-{\tan }^{2}A}-\frac{1}{\cos A}$

$\frac{\sec A+\tan A}{1}-\frac{1}{\cos A}$

$=\tan A$

$RHS\frac{1}{\cos A}-\frac{1}{\sec A+\tan A}+\frac{\sec A-\tan A}{\sec A-\tan A}$

$\frac{1}{\cos A}-\frac{\sec A-\tan A}{{\sec }^{2}A-{\tan }^{2}A}$

$\frac{1}{\cos A}-\sec A-\tan A$

$=\tan A$

$\therefore LHS=RHS$