Question

Consider this equation.

sin(θ)= 3sqrt10/10


If θ is an angle in quadrant II, what is the value of tan(θ) ?

Answer 1

tan x = -3 (Answer A)

Explanation:

We want to find the tangent of this angle "theta," and recall the trig identity

(sin x)^2 + (cos x)^2 = 1.

3√10

If sin x = -----------

10

90

then (sin x)^2 = ----------- = 9/10

100

and (cos x)^2 = 1 - 9/10 = 1/10

sin x 3√10/10

Then tan x = ---------- = -------------- = -3 (Answer A)

cos x 1√10/10

The tangent function is negative in Quadrant II. In Quadrant I tan x = +3

Answer 2

A

Explanation:

Using the trig identity

sin²x + cos²x = 1 , then cos x = ±
`\sqrt{1-((3√(10) )/(10))^2 }`

Given

sinθ =
`(3√(10) )/(10)` , then

cosθ = ±
`\sqrt{1-((3√(10) )/(10))^2 }`

= ±
`\sqrt{1-(9)/(10) }`

= ±
`\sqrt{(1)/(10) }`

Since θ is in second quadrant where cosθ < 0 , then

cosθ = -
`(1)/(√(10) )`

Then

tanθ =
`(sin0)/(cos0)` =
`((3√(10) )/(10) )/((-1)/(√(10) ) )` = - 3 → A