Given:


To find:
The quadrant of the terminal side of
and find the value of
.
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,


Here cos is positive and sine is negative. So,
must be lies in Quadrant IV.
We know that,



It is only negative because
lies in Quadrant IV. So,

After substituting
, we get





Therefore, the correct option is B.
C I think because the lines intercept at a common point
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Step-by-step explanation:
So, the function is obviously shifted up by 1 so it rules out A and B, and we also see that the period is 4pi rather than 2pi, and using the fact that:

Now that we know our b value, we see that the correct answer would be C.
Answer:
y =
x + 9
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Calculate m using the slope formula
m = 
with (x₁, y₁ ) = (- 2, 6 ) and (x₂, y₂ ) = (2, 12 )
m =
=
=
=
, then
y =
x + c ← is the partial equation
To find c substitute either of the 2 points into the partial equation.
Using (2, 12 ) , then
12 = 3 + c ⇒ c = 12 - 3 = 9
y =
x + 9 ← equation of line