Question

Sec A+Tan A =1/3,What is Sec A – Tan A ?

Answer 1

[tex]sec^{2} x- tan^{2} x=1 let secx-tanx be x sec^{2} x- tan^{2} x=1 [/tex]
(secx+tanx)(secx-tanx)=1
since $a^{2} - b^{2} = (a+b)(a-b)$
1/3(secx-tanx)=1
secx-tanx=3

Answer 2

We know that,
1+tan²A = sec²A
⇒sec²A - tan²A = 1
∵(a+b)(a-b) =(a²-b²)
∴sec²A - tan²A = 1
⇒(secA+tanA)(secA-tanA) =1
⇒(secA-tanA)×(1/3) =1
⇒(secA - tanA) = 3