Respuesta :

EEPJR
0
-1
-inf
1
0
0

repectively

Answer:

Step-by-step explanation:

We have to find the values of the given trigonometric ratios at the angle indicated. Thus,

(A) The given trigonometric function is:

[tex]cos270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]cos270^{\circ}=-(0)=0[/tex]

(B) The given trigonometric function is:

[tex]sin270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]sin270^{\circ}=-1[/tex]

(C) The given trigonometric function is:

[tex]tan270^{\circ}[/tex]

Since, the function lies in Quadrant III, therefore the value of the function will be negative.

Also, [tex]tan270^{\circ}=undefined[/tex]

(D) The given function is:

[tex]cos0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also,  [tex]cos0^{\circ}=1[/tex]

(E) The given function is:

[tex]sin0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also,  [tex]sin0^{\circ}=0[/tex]

(F) The given function is:

[tex]tan0^{\circ}[/tex]

Since, the function lies in Quadrant I, therefore the value of the function will be positive.

Also,  [tex]tan0^{\circ}=0[/tex]