Given the following right triangle, find cosθ, sinθ, tanθ, secθ, cscθ, and cotθ. Do not approximate: Find exact answers. Show all of your work and explain steps as necessary. Indicate your response here. Be sure to show all of your work and explain each step, as necessary. If you need help using the equation editor in Microsoft Word, refer to the Equation Editor Tutorial document.
Remember to address each of the critical elements of the prompt:
Establish a context for the problem by explaining in your own words the course principles that apply: What are the relationships between theta and the lengths of the sides of the triangle? Be sure to correctly use the appropriate terminology in your explanation.
Apply the mathematical process to solve the problem:
Use the Pythagorean theorem to find the third side of the triangle.
Write out the six trigonometric functions in exact form related to theta.
Clearly state the answer using appropriate precalculus notations.

See below for explanation

Step-by-step explanation:

A related question can be found at chegg. Find attached the diagram.

The triangle is a right angled triangle. The 3 sides are named opposite, adjacent and hypotenuse.

The opposite is the vertical side facing angle theta (θ)

The adjacent is horizontal side

The angle between opposite and adjacent = 90°

The hypotenuse is the longest side

The relationship between the 3 sides:

Hypotenuse² = opposite ² + adjacent²

The relationship between theta (θ)

and the lengths of the sides of the triangle:

Applying SOHCAHTOA in trigonometry

Sine ratio:

Sinθ = opposite/hypotenuse

Cosine ratio:

Cosθ = adjacent/hypotenuse

Tangent ratio:

Tanθ = opposite/adjacent

Applying the Pythagorean theorem to find the third side of the triangle

Hypotenuse² = opposite ² + adjacent²

Hypotenuse = 7, opposite = 4

adjacent² = Hypotenuse² - opposite ²

adjacent² = 7²-4² = 49-16

adjacent² = 33

Adjacent = √33

Write out the six trigonometric functions in exact form related to theta.

cosθ = adjacent/hypotenuse = (√33)/7

sinθ = opposite/hypotenuse = 4/7

tanθ = sinθ/cosθ = opposite/adjacent

sinθ/cosθ = 4/7 ÷ (√33)/7

= 4/7 × 7/(√33) = 4/(√33)

opposite/adjacent = 4/(√33)

tanθ = 4/(√33)

secθ = 1/cosθ = 1/(√33)/7

= 7/√33

cosecθ = 1/sinθ = 1/(4/7)

= 1×7/4 = 7/4

cotθ = 1/tanθ = 1/[4/(√33)]

cotθ =(√33)/4

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2