Sec A+Tan A =1/3,What is Sec A – Tan A ?
Answers
Answered by
2
SecA+tanA=1/3
//as we know....sec^2A-tan^2A=1
//(SecA-tanA)(SecA+tanA)=1
//(secA-tanA)(1/3)=1
//SecA-tanA=3
ansss:
Hope it helps
Really impressive !
Answered by
1
we know the trigonometric identity: sec² A- Tan²A =1
this can be written as (secA - tanA)(secA +tanA) =1 since a²+b² = (a+b)(a-b)
let this be eq(1)
given, (secA+tanA)=1/3
by substituting in the above eq(1)
we get,
(1/3)(secA-tanA) =1
secA -tanA = 1/(1/3)
⇒sec A - tan A = 3
this can be written as (secA - tanA)(secA +tanA) =1 since a²+b² = (a+b)(a-b)
let this be eq(1)
given, (secA+tanA)=1/3
by substituting in the above eq(1)
we get,
(1/3)(secA-tanA) =1
secA -tanA = 1/(1/3)
⇒sec A - tan A = 3
Similar questions