Question

if cos(θ)=10/26, 0≤θ≤π/2 determine the following. Leave results as exact values.

sin(θ)=
tan(θ)=
sec(θ)=

Answer 1

Draw the triangle in QI. It is your standard 5-12-13 triangle, scaled up, with
x = 10
r = 26
y = 24
sinθ = y/r
tanθ = y/x
secθ = 1/cosθ

Answer 2

sketch your triangle,
cosθ = 10/23 = 5/13 , since the cosine is + in I or IV, θ is in I or IV
either recognize the 5-12-13 triangle or use Pythagoras
x = 5, y= 6, r = 13

sinθ = 12/13 or sinθ = -12/13
tanθ = 6/5 or tanθ = -6/5
secθ = 1/cosθ = 13/5

Answer 3

my bad, I read the domain as
0 ≤ θ ≤ 2π , need to clean my glasses

sinθ = 12/13
tanθ = 6/5
secθ = 1/cosθ = 13/5

Answer 4

First simplify value of cos θ

cos θ = 10 / 26 = 2 ∙ 5 / 2 ∙ 13

cos θ = 5 / 13


sin θ = ± √ ( 1 - cos² θ )

0 ≤ θ ≤ π / 2 is Quadrant I

In Quadrant I all trigonometric functions are positive so:

sin θ = √ ( 1 - cos² θ )

sin θ = √ [ 1 - ( 5 / 13 )² ]

sin θ = √ ( 1 - 25 / 169 )

sin θ = √ ( 169 / 169 - 25 / 169 )

sin θ = √ ( 144 / 169 )

sin θ = √144 / √169

sin θ = 12 / 13


tan θ = sin θ / cos θ = ( 12 / 13 ) / ( 5 / 13 ) = 13 ∙ 12 / 5 ∙ 13 = 12 / 5


sec θ = 1 / cos θ = 1 / ( 5 / 13 ) = 13 / 5