Question

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

Answer 1

To prove:

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

Considering LHS

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

$=\frac{\sec A-\tan A+\sec A+\tan A}{(\sec A+\tan A)(\sec A-\tan A)}$

$=\frac{2\sec A}{{\sec }^{2}A-{\tan }^{2}A}$

$=2\sec A$ As $({\sec }^{2}A-{\tan }^{2}A=1)$

Hence the answer is $=2\sec A$