In which of the following intervals does the trigonometric inequality sec(x) < cot(x) always hold true.
1 answer:
The inequality for sec (x) < cot (x) is; π/2 < x < π
<h3>How to express trigonometric inequality?</h3>
We are given that;
We want to find the intervals that the trigonometric inequality sec (x) < cot (x) always hold true.
This can also be expressed as;
1/cos (x) < 1/tan (x)
Now, this can happen only in the quadrant where tan (x) is negative and cos x is positive which is in fourth quadrant where;
π/2 < x < π
The inequality for sec (x) < cot (x) is; π/2 < x < π
Read more about Trigonometric Inequality at; brainly.com/question/12094532
#SPJ1
You might be interested in
Answer:
equivalent to 
Step-by-step explanation:
Given ratio,
It can be written as 
cancel common factor,
Answer:
385.6
Step-by-step explanation:
The wage times the hours worked find the total she makes.
Their profit made will be 4,420
This is because 0.60 x 1300 is 780
And the amount they will make by selling is $5200, because 1300 x 4
So we will do 5200 - 780, and that is the profit made- $4,420
Answer:
39ft²
Step-by-step explanation:
Divide up the Shape
Big rectangle:
a=lw
11*3=33ft
Small rectangle
2*3=6ft
Whole shape:
33+6=39ft²
The answer is d because 12c=20 = c=65/12 divide by both sides
