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.