# If $\sec x\tan x<0$, then must we have $\tan x<0$? $\csc x \cot x < 0$? that $x$ is in the third or fourth quadrant?

I had some trouble solving this problem from Barron's Math Level 2 workbook.

If $$(\sec x)(\tan x) < 0$$, which of the following must be true?

I. $$\tan x < 0$$

II. $$\csc x \cot x < 0$$

III. $$x$$ is in the third or fourth quadrant

I immediately eliminated Statement I, because either one of the factors, or both of them, could be less than 0, but the question asks for which statements must be true.

Statement II, I figured was wrong because the reciprocal of $$\sec x \tan x$$ is $$\cos x \cot x$$, not $$\csc x \cot x$$.

Statement III, however, I did not understand how to definitively prove or disprove. I converted the given expression to $$\frac{\sin x}{\cos^2x}$$. For it to be less than 0, one of the numerator or denominator must be less than 0. $$\sin x$$ is positive in Quadrants I and II, so it is negative in III and IV. $$\cos^2x$$ cannot be negative, so it must be in Quadrants I or IV, where it is positive. Therefore, $$\sin x$$ must be in Quadrants III and IV (to be negative) and $$\cos x$$ in Quadrants I or IV (to be positive). Where do I take it from there? The text says choice III is true, only.

• Part III is determined only by the sign of $\sin x$ because $\cos^2 x$ is nonnegative for all $x$ (due to the squaring) – MPW Oct 27 at 18:16

Doing it as you did, you know that $$\cos^2x$$ is always positive. Therefore, the expression is negative if and only if $$\sin x$$ is negative. As you note, that happens only in quadrants III and IV, so the third statement is true.
• Another question: For this problem, is it safe to assume that $\cos^2 x \ne 0$? – bin Oct 27 at 18:32