Find the values of the six trigonometric functions of theta based on the following constraints:
1) Theta is found in quadrant II.
2) sin theta = 3/5

Solution:

Given, θ lies in second quadrant

sin θ = 3/5

We have to find the values of the six trigonometric functions of theta.

Using trigonometric ratios,

sin θ = opposite/hypotenuse

So, opposite = 3

Hypotenuse = 5

Using Pythagorean theorem,

(Hypotenuse)² = (Opposite)² + (Adjacent)²

(5)² = (3)² + (Adjacent)²

25 = 9 + (Adjacent)²

(Adjacent)² = 25 - 9

(Adjacent)² = 16

Taking square root,

Adjacent = 4

In second quadrant, angle is (180° - θ)

Sine is positive

Cosine is negative

Tangent is negative

Cosecant is positive

Secant is negative

Cotangent is negative

We know, cos θ = adjacent/hypotenuse

So, cos(180° - θ) = - cos θ

cos θ = -4/5

We know, tan θ = opposite/adjacent

So, tan(180° - θ) = - tan θ

tan θ = -3/4

We know, cosec θ = 1/sin θ

So, cosec(180° - θ) = cosec θ

cosec θ = 1/(3/5)

cosec θ = 5/3

We know, sec θ = 1/cos θ

So, sec(180° - θ) = - sec θ

sec θ = 1/(-4/5)

sec θ = -5/4

We know, cot θ = 1/tan θ

So, cot(180° - θ) = - cot θ

cot θ = 1/(-3/4)

cot θ = -4/3

Therefore, the values of six trigonometric functions are sin θ = 3/5, cos θ = -4/5, tan θ = -3/4, cosec θ = 5/3, sec θ = -5/4 and cot θ = -4/3.

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Find the values of the six trigonometric functions of theta based on the following constraints: 1) Theta is found in quadrant II. 2) sin theta = 3/5

Summary:

The values of the six trigonometric functions of theta based on the following constraints: 1) Theta is found in quadrant II, 2) sin theta = 3/5 are sin θ = 3/5, cos θ = -4/5, tan θ = -3/4, cosec θ = 5/3, sec θ = -5/4 and cot θ = -4/3.