Domain of A Composite Trig Function

[darkchild]

Homework Statement

Suppose that the function f is defined on an interval by the formula
f(x) = \sqrt{tan^{2}x - 1}. If f is continuous, which of the following intervals could be its domain?
(A) (\frac{3\pi}{4},\pi)

(B) (\frac{\pi}{4},\frac{\pi}{2})

(C) (\frac{\pi}{4},\frac{3\pi}{4})

(D) (-\frac{\pi}{4},0)

(E) (- \frac{3\pi}{4},- \frac{\pi}{4})

The correct answer is supposed to be B.

Homework Equations

none

The Attempt at a Solution

cos(x) can't be zero, so that rules out choices (C) and (E). sin(x) cannot be zero.

This gave me an idea about the range of possible values: (excluding x=0, of course)
tan^{2}x - 1 \geq 0

tan^{2}x \geq 1

-1 \leq tan(x) \leq 1

- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}

I can't figure out how to eliminate choices A and D. The three remaining choices all seem to me to be correct.

 

[Mark44]

Homework Statement

Suppose that the function f is defined on an interval by the formula
f(x) = \sqrt{tan^{2}x - 1}. If f is continuous, which of the following intervals could be its domain?
(A) (\frac{3\pi}{4},\pi)

(B) (\frac{\pi}{4},\frac{\pi}{2})

(C) (\frac{\pi}{4},\frac{3\pi}{4})

(D) (-\frac{\pi}{4},0)

(E) (- \frac{3\pi}{4},- \frac{\pi}{4})

The correct answer is supposed to be B.

Homework Equations

none

The Attempt at a Solution

cos(x) can't be zero, so that rules out choices (C) and (E). sin(x) cannot be zero.

This gave me an idea about the range of possible values: (excluding x=0, of course)
tan^{2}x - 1 \geq 0

tan^{2}x \geq 1

-1 \leq tan(x) \leq 1

Mistake on the line above. If the inequality above this line were tan²(x) <= 1, then this would be correct.

For tan²(x) >= 1, you have either tan(x) >= 1 or tan(x) <= -1.

- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}

I can't figure out how to eliminate choices A and D. The three remaining choices all seem to me to be correct.