Question

RootsecA-1/secA+1 = root secA+1/secA-1 = 2cosec

Answer 1

√(secA-1/secA+1)+√(secA+1/secA-1)
=√{(secA-1)(secA-1)/(secA+1)(secA-1)}+√{(secA+1)(secA+1)/(secA-1)(secA+1)}
=√{(secA-1)²/(sec²A-1)}+√{(secA+1)²/(sec²A-1)}
=√(secA-1)²/tan²A+√(secA+1)²/tan²A
=(secA-1)/tanA+(secA+1)/tanA
=secA/tanA-1/tanA+secA/tanA+1/tanA
=(1/cosA)/(sinA/cosA)-cotA+(1/cosA)/(sinA/cosA)+cotA
=1/sinA+1/sinA
=2/sinA
=2cosecA (Proved)

Answer 2

This is the proper solution please leave a comment if you dont understand