Question

Sec A +Tan A = 5 , Sec A =

Answer 1

Answer:

= > secA + tanA = 5

Square on both sides:

= > (secA+tanA)^2 = 5^2

= > sec^2A + tan^2 A + 2secAtanA = 25

We know, tan^2 A = sec^2A - 1

= > sec^2 A + sec^2 A - 1 + 2secAtanA = 25

= > 2sec^2A + 2secAtanA = 26

= > sec^2A + secAtanA = 13

= > sec^2A - 13 = secAtanA

Square on both sides,

= > sec^4A + 169 - 26sec^2A = sec^2Atan^2A

= > sec^4A + 169 - 26sec^2A = sec^2A(sec^2A-1)

Let sec^2A = a

= > a^2 + 169 - 26a = a^2 - a

= > 169 = 26a - a

= > 169 = 25a

= > 169/25 = a

= > 169/25 = sec^2A

= > $\pm$13/5 = secA

Required value of secA is $\pm$ 13/5