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

Given, $\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A+\tan A}$
or $\frac{1}{\sec A-\tan A}+\frac{1}{\sec A+\tan A}=\frac{1}{\cos A}+\frac{1}{\cos A}$
Here, R.H.S.$=\frac{2}{\cos A}$
Now,
L.H.S.$=\frac{1}{\sec A-\tan A}+\frac{1}{\sec A+\tan A}=\frac{\sec A+\tan A+\sec A-\tan A}{(\sec A-\tan A)(\sec A+\tan A)}=\frac{2}{\cos A}$
Thus, L.H.S.=R.H.S.