Question

sec x + tan x=9, find cos x

Answer 1

we know that,
${sec}^{2} x - {tan}^{2} x = 1$
which can be separated as
$(sec x - tan x)(sec x + tan x) = 1$
sec X - tan X= 1/9 ------------ eqn 1
sec X + tan X=9 ------------eqn 2
on adding eqn 1 and eqn 2
2 sec X = 82/9
sec X = 41/9
therefore, cos x = 9/41

Answer 2

sec(x) is related to cos(x) in the relation

sec(x) = 1/cos(x)

You are given tan(x) = 3 and cos(x)> 0

 Note:

1st Quadrant : sin(x) > 0, cos(x) > 0, and tan(x) > 0

2nd Quadrant: sin(x) < 0, cos(x) > 0, and tan(x) < 0

3rd Quadrant: sin(x) > 0, cos(x) < 0, and tan(x)  < 0

4th Quadrant: sin(x) < 0, cos(x) < 0, and tan(x) > 0

 

since both cos(x) > 0, and tan(x) > 0 this is the first quadrant

tan(x) = opp/adj = 3/1 , therefore opp = 3, adj = 1, find hyp because cos(x) = adj/hyp. To do this use the Pythagorean theorem

hyp2  =opp2 + adj2

hyp2  = 32 + 12 = 9 +1 = 10

hyp = ±√10 we use the positive value since  cos(x) > 0

cos(x) = adj/hyp

          =1/√10

sex(x) = 1/cos(x) = 1/1/√10 =√10

 

sec(x) = √10 and cos(x) =1/√10