Question

Prove :

$\sqrt{ \frac{secA -1}{secA +1} } + \sqrt{ \frac{secA + 1}{secA -1} } = 2cosecA$

Answer 1

You just have to rationalise each of the square roots 
$\sqrt{ \frac{(secA-1)(secA-1)}{sec^2A-1} } + \sqrt{ \frac{(secA+1)(secA+1)}{sec^2A-1} }$
[tex] \\ \frac{sec A -1 }{tan A} + \frac{sec A +1 }{tan A} = \frac{2sec A}{tan A } =2cosec A[/tex]