Question

Prove that: $\frac{1}{\sec A-1}+\frac{1}{\sec A+1}=2\csc A\cot A$

Answer 1

L.H.S

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

$=\frac{\sec A+1+\sec A-1}{(\sec A-1)(\sec A+1)}$

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

$=\frac{2\sec A}{{\tan }^{2}A}[\because {\sec }^{2}A-1={\tan }^{2}A]$

$=\frac{\frac{2}{\cos A}}{\frac{{\sin }^{2}A}{{\cos }^{2}A}}$

$=\frac{2\cos A}{{\sin }^{2}A}$

$=2\csc A\cot A$

Hence, this is the answer.