Question

Please Help! In which of the following intervals does the trigonometric inequality sec(x)

Answer 1

• secx=1/cosx
• cotx=cosx/sinx

0 is included in first one it can't be a choice

Test second one

• secπ/2<cotπ/2
• secπ<cotπ

Hence it's true

Third one

• secπ<cotπ
• sec270<cot270

This is also true

Option B and C are correct

Answer 2

Since this is multiple choice, we can pick any x from the listed intervals and check whether sec(x) < cot(x) is true.

• From 0 < x < π/2, take x = π/4. Then

sec(π/4) = 1/cos(π/4) = √2

cot(π/4) = 1/tan(π/4) = 1

but √2 < 1 is not true.

• From π/2 < x < π, take x = 3π/4. Then

sec(3π/4) = 1/cos(3π/4) = -√2

cot(3π/4) = 1/tan(3π/4) = -1

and -√2 < -1 is true. So sec(x) < cot(x) is always true for π/2 < x < π.

• From π < x < 3π/2, take x = 5π/4. Then

sec(5π/4) = 1/cos(5π/4) = -√2

cot(5π/4) = 1/tan(5π/4) = 1

and -√2 < 1 is true. So sec(x) < cot(x) is also true for π < x < 3π/2.

• From 3π/2 < x < 2π, take x = 7π/4. Then

sec(7π/4) = 1/cos(7π/4) = √2

cot(7π/4) = 1/tan(7π/4) = -1

but √2 < -1 is not true.